{ "query": "Summarize in 5 sentences the contents of your context. Speak in Old English manner similar to how Shakespeare writes his sonnets.", "namespace": "cdf5e841-bd92-4996-b7f2-7c2a398184c5", "messages": [ { "role": "user", "content": "Summarize in 5 sentences the contents of your context. Speak in Old English manner similar to how Shakespeare writes his sonnets." }, { "role": "assistant", "content": " " } ], "stream": true, "language_level": "", "chat_channel": ":160:1055", "language": "German", "tone": "neutral", "writing_style": "standard", "model": "gpt-4o-mini-2024-07-18", "knowledgebase": "ki-dev-large", "seed": null, "client_id": 1055, "all_context": true, "follow_up_for": null, "knowledgebase_files_count": 2, "override_command": "", "disable_clarity_check": false, "custom_primer": "", "logging": true, "query_route": "" } INITIALIZATION Knowledgebase: ki-dev-large Base Query: Summarize in 5 sentences the contents of your context. Speak in Old English manner similar to how Shakespeare writes his sonnets. Model: gpt-4o-mini-2024-07-18 Use Curl?: ================================================== **Elapsed Time: 0.00 seconds** ================================================== ROUTING Query type: summary ================================================== **Elapsed Time: 1.59 seconds** ================================================== PRIMER Primer: You are Simon, a highly intelligent personal assistant in a system called KIOS. You are a chatbot that can read knowledgebases through the "CONTEXT" that is included in the user's chat message. Your role is to act as an expert at summarization and analysis. In your responses to enterprise users, prioritize clarity, trustworthiness, and appropriate formality. Be honest by admitting when a topic falls outside your scope of knowledge, and suggest alternative avenues for obtaining information when necessary. Make effective use of chat history to avoid redundancy and enhance response relevance, continuously adapting to integrate all necessary details in your interactions. Use as much tokens as possible to provide a detailed response. ================================================== **Elapsed Time: 0.37 seconds** ================================================== FINAL QUERY Final Query: CONTEXT: ########## File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 1 Context: # TESTED BY: Lois Layne Ramos ## TEST EMAIL: loislayneramos@gmail.com ## WEB APP URL: [https://tester-task.vercel.app/](https://tester-task.vercel.app/) ## BROWSER: Chrome, Safari, Edge, Firefox, Opera ## OS: Mac OS, Windows ## DATE OF TESTING: 02-13-24 # BUGS/ISSUES FOUND: 1. **[FUNCTIONAL]** INPUT FIELD: Accepts empty values 2. **[FUNCTIONAL]** KEYBOARD FUNCTION: Click 'ENTER' KEY with/without input values create double entries. 3. **[USER INTERFACE GLITCHES]** DELETE BUTTON: Delete all data in one click - no confirmation asked. 4. **[USER INTERFACE GLITCHES]** BUTTON LABEL: Function is unclear. 5. **[USER INTERFACE GLITCHES]** DUPLICATE VALUES: 2 entries added with the same input values. 6. **[SECURITY VULNERABILITIES]** LOGIN: No login page - data is not secure. # PROBLEM DESCRIPTION The user is still able to create/add tasks even without any input. Allowing this might result in a waste of space and an overwhelming number of empty entries. (Refer to bug ticket no. 1) #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 3 Context: # PROBLEM DESCRIPTION Pressing the Enter key (regardless of whether there's data entered) results in 2 entries being created: an empty entry and the user's input value. (Refer to bug ticket no. 2) ## Screenshot: ![TODO APP](https://via.placeholder.com/300) ## Steps to replicate 1. Go to [tester-task.vercel.app](https://tester-task.vercel.app/) 2. Click input field 3. Leave textarea blank/ with input value 4. Press 'Enter' in keyboard Image Analysis: ### Comprehensive Examination #### 1. **Localization and Attribution:** - **Image 1:** - Positioned below the text "Screenshot:" and above the "Steps to replicate." - Assigned number: **Image 1**. #### 2. **Object Detection and Classification:** - **Image 1:** - **Objects Detected:** - A digital interface of a "TODO APP." - Text elements such as "Tasks," "Done Tasks," and "TASKS." - Input areas labeled with placeholder text like "New task." - Icons including checkmarks, a plus icon, and a red delete icon. - **Key Features:** - The input fields with checkboxes, indicating they may be tasks or items to be marked as complete. - A distinct blue "Tasks" tab with a number (10) indicating possibly the number of ongoing tasks. - A red "TASKS" tab likely for accessing task functionalities. #### 3. **Scene and Activity Analysis:** - **Image 1:** - **Scene Description:** - The scene showcases a task management interface of a TODO app. - The background is blue, providing a contrast to the primarily white input areas. - **Activities Taking Place:** - Users are presented with a list of tasks ("test 9," "24") and are able to add new tasks. - There are functional components to mark tasks as complete or delete them. #### 4. **Text Analysis:** - **Extracted Text:** - "TODO APP," "Tasks," "Done Tasks," "TASKS," "test 9," "24," "New task." - **Text Content Analysis:** - The text explains the structure and purpose of the app. - "Tasks" tab indicates the total number of tasks (10). - "Done Tasks" likely denotes completed tasks, although the exact number is not visible. - Placeholder and task names give a quick insight into the app's functionality. #### 7. **Anomaly Detection:** - **Image 1:** - **Possible Anomalies:** - The text description highlights a bug where pressing the Enter key creates two entries (an empty entry and the user’s input). - **Impact:** - This bug impacts the usability of the app, leading to unwanted empty entries and potentially cluttered task lists. #### 8. **Color Analysis:** - **Image 1:** - **Dominant Colors:** - Blue: Background, conveying a calm and professional feel. - White: Main input areas, enhancing readability. - Red: Delete icons, drawing attention to functions for removing tasks. - **Impact on Perception:** - The color scheme ensures that key elements like tasks and interactive icons stand out clearly. #### 9. **Perspective and Composition:** - **Perspective:** - Front-facing view of a user interface on a digital screen. - **Composition:** - The interface is composed centrally with task lists and interactive elements balanced within the screen boundaries. - Clear segregation of task categories with respective tabs. #### 10. **Contextual Significance:** - **Context:** - This image is within a bug report or a similar documentation context. - **Contribution:** - The image visually supports the written problem description by illustrating the issue within the TODO app. #### 12. **Graph and Trend Analysis:** - **Process Description (Depicted through steps):** - Specifies steps to replicate the described bug: 1. Visit the given URL. 2. Click the input field. 3. Leave it blank or enter a value. 4. Press 'Enter'. - **Significance:** - Ensures precise replication for developers or testers to observe and address the issue. #### 13. **Ablaufprozesse (Process Flows):** - The described steps detail a straightforward process flow to replicate a bug, critical for identifying and resolving the issue reported. This examination spans various aspects, helping in understanding the content, functionality, and issues of the depicted TODO app. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 4 Context: # Actual Behavior - Clicking the "Enter" key, regardless of whether any data is entered, results in the creation of two entries instead of one. # Expected Behavior - Pressing the Enter key should only create a single new task entry, regardless of whether any information is currently displayed in the input field. # PROBLEM DESCRIPTION 3. The "Delete All Tasks" button lacks a confirmation prompt, increasing the risk of accidental data loss. This was evident during testing when a user unintentionally deleted all tasks due to the absence of a confirmation step. (Refer to bug ticket no. 3) # Screenshot ![Screenshot]() Image Analysis: ### Comprehensive Examination: #### 1. Localization and Attribution: - **Image Numbering:** - Image 1: Located at the top, containing textual content. - Image 2: Located at the bottom, depicting a screenshot of a digital interface. #### 2. Object Detection and Classification: - **Image 2:** - Objects: - A digital interface (a To-Do App). - Three task categories: "Tasks," "Done Tasks," and "DELETE ALL TASKS." - A text input field labeled "New task." - A plus button next to the text input field. - Circular progress indicators or buttons next to "Tasks" and "Done Tasks" indicating status. - A notification dot adjacent to "TASKS." #### 3. Scene and Activity Analysis: - **Image 2:** - Scene: A digital To-Do application interface. - Activities: - Users can add a new task via an input field. - Tasks can be categorized or completed under "Tasks" or "Done Tasks." - There is an option to delete all tasks without an apparent confirmation prompt. #### 4. Text Analysis: - **Image 1:** - Actual Behavior: Clicking the "Enter" key, regardless of whether any data is entered, results in the creation of two entries instead of one. - Expected Behavior: Pressing the Enter key should only create a single new task entry, regardless of whether any information is currently displayed in the input field. - Problem Description: The "Delete All Tasks" button lacks a confirmation prompt, increasing the risk of accidental data loss. This was evident during testing when a user unintentionally deleted all tasks due to the absence of a confirmation step. - **Image 2:** - Text on screen: - "TODO APP" - "Tasks" - "Done Tasks" - "TASKS" - "No data found" - "New task" #### 6. Product Analysis: - **Image 2:** - A To-Do application depicted in the screenshot. - Key Features: Task listing, task status indicators, input field for new tasks, and a prominent delete button. - Materials and Colors: Predominantly minimalistic UI with a white input box, blue progress state for "Done Tasks," and a red delete button. #### 7. Anomaly Detection: - **Image 2:** - **Anomalies:** - The "Delete All Tasks" button has a prominent position and coloration without an evident confirmation step, which could lead to accidental data loss. - The functionality glitch creating two entries instead of one with the "Enter" key mentioned in the textual content. #### 8. Color Analysis: - **Image 2:** - Dominant Colors: Blue for the main theme, red for the delete action, and white for the input areas. - Impact: The use of red color for the delete button draws immediate attention to it, highlighting its importance and potential risk. #### 9. Perspective and Composition: - **Image 2:** - Perspective: The screenshot appears to be taken from a straight-on view. - Composition: The elements are neatly arranged with the input area at the bottom, task categories at the top, and status indicators noticeable. #### 10. Contextual Significance: - **Overall:** - The images contribute to understanding the usability issues and functional glitches in the To-Do application. The textual content contextualizes the problem with the application's usage, emphasizing the necessity for better user interaction handling and safeguards against accidental data loss. #### 12. Graph and Trend Analysis: - **Image 1:** - The problem description and behavior analysis indicate trends associated with the issues faced by users while interacting with the To-Do application, particularly focusing on task entry duplication and deletion risks. #### 14. Ablaufprozesse (Process Flows): - **Image 2:** - Process Flow: User interaction involves adding new tasks, categorizing/completing them, and potentially deleting all tasks. The flow could lead to accidental data loss in the absence of a confirmation step for the "Delete All Tasks" button. #### 15. Prozessbeschreibungen (Process Descriptions): - **Image 1:** - Describes the process and expected vs. actual behavior for task entry and deletion in the To-Do app interface. ### Conclusion: These analyses point out key usability concerns and the need for design improvements to avoid user errors in the To-Do application. Both the textual explanations and the visual representation help in understanding the impact of these issues on user experience. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 5 Context: # Steps to replicate 1. Go to [https://tester-task.vercel.app/](https://tester-task.vercel.app/) 2. Create at least 20 entries. 3. Click the delete task button. ## Actual Behavior - Clicking the "Delete Task" button results in the immediate deletion of all tasks without requiring confirmation. ## Expected Behavior - Clicking the "Delete Task" button should prompt the user for confirmation before permanently removing any data. ## Problem Description The design of the "Done Tasks" button is confusing due to a conflicting element. Upon clicking a specific entry inside the table, a second button with an "x" appears, suggesting task removal, despite the original button label implying task completion. (Refer to bug ticket no. 4) ## Screenshot ![Screenshot](screenshot_link) Image Analysis: ### Examination of the Attached Visual Content #### Image Localization and Attribution - **Image 1**: The text content and instructional steps. - **Image 2**: The screenshot of the "TODO APP". #### Object Detection and Classification - **Image 1**: - Objects: Text blocks. - **Image 2**: - Objects: User interface of a TODO application, including buttons, text fields, and a table. #### Scene and Activity Analysis - **Image 1**: - Scene: An instructional and descriptive document detailing a procedure for a "TODO APP". - Activities: Steps to replicate a task deletion issue, description of actual and expected behavior, problem description. - **Image 2**: - Scene: A screenshot of the "TODO APP" interface. - Activities: Interaction with tasks within the app such as adding or completing tasks. #### Text Analysis - **Image 1**: - **Steps to replicate**: 1. Go to https://tester-task.vercel.app/ 2. Create at least 20 entries. 3. Click delete task button - **Actual Behavior**: - Clicking the "Delete Task" button results in the immediate deletion of all tasks without requiring confirmation. - **Expected Behavior**: - Clicking the "Delete Task" button should prompt the user for confirmation before permanently removing any data. - **Problem Description**: - The design of the "Done Tasks" button is confusing due to a conflicting element. Upon clicking a specific entry inside the table, a second button with an "x" appears, suggesting task removal, despite the original button label implying task completion. - Reference: Bug ticket no. 4. - **Screenshot**: - Caption indicating a visual representation of the problem described. - **Image 2**: - Contains labels like "Tasks", "Done Tasks", "T+STS", "New task", and an "x" button indicating deletion. #### Color Analysis - **Image 2**: - Dominant Colors: Blue background. - Other Colors: White for text fields and buttons, red for the delete button, and gray for the text entries. - Impact: The color composition highlights the interactive elements, particularly the red delete button, drawing attention to it. #### Perspective and Composition - **Image 2**: - Perspective: Frontal view of the application interface. - Composition: The interface is centrally aligned with clearly delineated sections for tasks, done tasks, and task management buttons, providing a clean and organized visual structure. ### Summary The document contains detailed steps to reproduce a bug in a TODO application, describes the actual and expected behaviors, and outlines a design issue related to the "Done Tasks" button that may confuse users. The included screenshot (Image 2) visually supports the problem description by highlighting the TODO APP interface and illustrating the confusing elements. The contextual and functional analysis intends to guide developers or testers in identifying and resolving the issues. The color, perspective, and composition analyses provide insight into how the interface design might impact user interaction. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 6 Context: # Steps to Replicate 1. Go to [https://tester-task.vercel.app/](https://tester-task.vercel.app/) 2. Click the “check” icon of the task 3. See 'x' button appear inside the ‘done tasks’ button ## Actual Behavior - An 'x' button appears inside the 'done tasks' button after selecting a task to mark as complete. ## Expected Behavior - The "Done Tasks" button should indicate its function, either by marking the selected task(s) as complete or by permanently removing them - this should be a separate button. ## Problem Description 5. Attempting to add two entries with identical input values currently results in duplicate entries being created without any notification or warning to the user. (Refer to bug ticket no. 5) #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 7 Context: # TODO App Screenshot ![TODO App Screenshot](https://tester-task.vercel.app/) ## Steps to replicate 1. Go to [tester-task.vercel.app](https://tester-task.vercel.app/) 2. Click the input field 3. First input value - `test1` 4. Click submit/add tasks button `(+)` 5. Second input value - `test1` 6. Click submit/add tasks button `(+)` ## Actual Behavior - The system currently allows users to add duplicate entries. Entering the same information twice results in two separate entries being created, instead of recognizing and preventing the duplication. Image Analysis: ### Comprehensive Examination of the Visual Content #### 1. Localization and Attribution: - **Image Identification and Location:** - **Image 1:** Positioned at the top of the page. - **Description:** Illustrates a section of a "TODO APP" interface. #### 2. Object Detection and Classification: - **Image 1:** - **Objects Detected:** - **Interface Elements:** - **Task List:** Includes several tasks labeled "test1" and "test 1." - **Buttons:** "Tasks," "Done Tasks," a red button labeled "Tasks," and checkbox icons next to each task. - **Key Features:** - **Tasks Tab:** Shows an active count of "22." - **Done Tasks Tab:** Shows an active count of "1." - **Task Addition Area:** Placeholder for task entries with checkboxes. #### 3. Scene and Activity Analysis: - **Image 1:** - **Scene Description:** - The image depicts a user interface of a TODO application. - **Activities Taking Place:** - The task list showcases duplicate entries of tasks, specifically labeled as "test1" and "test 1." #### 4. Text Analysis: - **Image 1:** - **Extracted Text:** - "TODO APP," "Tasks 22," "Done Tasks 1," "test1," "test 1" - **Significance:** - The text indicates task labels and numbers of tasks and done tasks, crucial for managing and tracking activities within the application. - **Text Below Image:** - **Steps to Replicate:** - Detailed instructions beginning with navigation to a URL, indicating input actions ("Click input field"), value entries ("First input value - test1"), and use of a submission button. - **Actual Behavior:** - Description of an observed issue where the system allows duplicate task entries, highlighting a potential bug. #### 8. Color Analysis: - **Image 1:** - **Color Composition:** - Dominant Colors: Light blue for the background of the application interface, white for the task entry areas and tabs, and red for the "Tasks" button. - **Impact on Perception:** - The color scheme provides a clear, user-friendly interface. The use of red for the "Tasks" button makes it stand out, likely indicating its importance or action button that requires attention. #### 9. Perspective and Composition: - **Image 1:** - **Perspective:** - Frontal view, typical for interface screenshots to give the viewer a direct look at how elements appear. - **Composition:** - Interface elements are arranged in a vertical scrollable list, with tabs at the top for navigation. The layout is clear and designed to be easily navigable. #### 10. Contextual Significance: - **Image 1:** - **Context within Document:** - The image is part of a bug reporting document, demonstrating an issue found within the TODO application interface. - **Contribution:** - Provides visual evidence of the described issue of duplicate task entries, supporting the written description and aiding in the understanding and replication of the bug. ### Summary: The visual content analyzed is a part of a user report highlighting a bug in a TODO application. The screenshot, textual instructions, and description combine to illustrate the issue of the application allowing duplicate task entries, providing a clear pathway for replication and addressing the bug. The use of clear, bright colors in the interface design enhances user experience by making navigation and interaction with the app straightforward and visually engaging. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 8 Context: # Expected Behavior - When a user tries to add an entry with duplicate values, the system should identify the duplicate: recognize that the input matches an existing entry and notify the user. # PROBLEM DESCRIPTION 6. The absence of a login page creates a security vulnerability. When a user opens the web application URL in a browser, they are directly transferred to the main page without authentication. This exposes sensitive user data to unauthorized access, as anyone with the URL can potentially view it (refer to bug ticket no. 6). # Screenshot ![Screenshot](https://tester-task.vercel.app) Image Analysis: ### Analysis of the Attached Visual Content #### 1. **Localization and Attribution** - **Text Content:** - Located at the top and mid sections of the page. - **Image:** - Located at the bottom section of the page. #### 2. **Object Detection and Classification** ##### Image 1: Screenshot of TODO App - **Objects Detected:** - Browser bar - Web application interface containing: - Header section - Task input field - Task list - Buttons for "Tasks," "Done Tasks," and "TASKS" - **Classified Categories:** - User Interface Elements - Text Fields - Buttons #### 3. **Scene and Activity Analysis** ##### Image 1: Screenshot of TODO App - **Scene Description:** - The image portrays a web application named "TODO APP". - **Activities Taking Place:** - A task management interface is shown, listing tasks named "test1" and "test 1" multiple times. - **Main Actors and Actions:** - The user seems to be managing tasks, with the interface displaying a list of tasks alongside options to add a new task and toggle between different views of tasks. #### 4. **Text Analysis** ##### Text Content: - **Expected Behavior:** - Description of how the system should behave when encountering duplicate values in input. - Text: "When a user tries to add an entry with duplicate values, the system should Identify the duplicate: Recognize that the input matches an existing entry - notify the user." - **Problem Description:** - Identifies security issue due to absence of a login page in a web application. - Text: "The absence of a login page creates a security vulnerability. When a user opens the web application URL in a browser, they are directly transferred to the main page without authentication. This exposes sensitive user data to unauthorized access, as anyone with the URL can potentially view it." - **Screenshot Text:** - "TODO APP" - Task entries: "test1", "test1", "test 1" - Buttons labeled: "Tasks," "Done Tasks," "TASKS" #### 6. **Product Analysis** ##### Image 1: Screenshot of TODO App - **Product Details:** - **Main Features:** - Interactive task management application. - Functional buttons for task filtering ("Tasks," "Done Tasks," "TASKS"). - Task input field for adding new tasks. - List of current tasks displayed. - **Materials:** - Digital interface with a clean and simple design. - **Colors:** - Dominant colors are white for the main app interface and blue for the background. #### 7. **Anomaly Detection** ##### Image 1: Screenshot of TODO App - **Anomalies:** - The security concern mentioned highlights a significant anomaly in web application design, as the lack of a login feature poses a security risk. #### 8. **Color Analysis** ##### Image 1: Screenshot of TODO App - **Color Composition:** - Dominant use of white for the application interface. - Blue gradient background for contrast. - Red button indicating active selection (likely for deletion or high priority). #### 9. **Perspective and Composition** ##### Image 1: Screenshot of TODO App - **Perspective:** - Standard front-view interface screenshot. - **Composition:** - The user interface is centered, with the browser bar at the top and main functional areas (task list, input field, and buttons) in the middle. #### 10. **Contextual Significance** - The image and the text combined highlight a critical issue and its expected behavior in managing tasks within a web application. It underscores the importance of secure authentication mechanisms in web applications to protect sensitive user data. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 9 Context: # Steps to replicate 1. Go to [https://tester-task.vercel.app/](https://tester-task.vercel.app/) 2. Redirect to main page ## Actual Behavior Accessing the web app URL directly redirects to the main page, bypassing the intended landing page or login process. ## Expected Behavior Upon accessing the web application, users should be prompted to log in before being granted access to the main page and their data. This ensures that only authorized users can view and interact with sensitive information. # Performance and Browser Compatibility Testing ## Results The web application demonstrated strong performance and consistent functionality across various browsers. No significant issues were identified during testing. ## Details - Data loading remained quick and responsive even when reaching 100 entries. - No performance degradation was observed when reloading the browser. - The application functioned identically on different browsers, including Opera, Safari, Chrome, Microsoft Edge, and Firefox. ## Conclusion These findings indicate a well-optimized application that delivers a reliable and consistent user experience across various performance scenarios and browser platforms. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 10 Context: # Additional Recommendations While no major issues were found, consider conducting further testing under simulated peak loads or with larger datasets to ensure continued performance stability. Monitor user feedback and performance metrics in real-world usage to identify any potential issues that may not have been captured during controlled testing. # Suggestions for Improvements 1. **Implement a hover text or tooltip for the "Delete All Tasks" button** that clearly explains its function. Introduce a confirmation dialog before deleting all tasks, requiring user confirmation and providing an option to cancel. (ref. Bug ticket no. 3) 2. **Implement a dedicated column** within the panel of task entries that clearly displays the total number of tasks created by the user. This would enhance user awareness and facilitate informed decision-making regarding their tasks. Furthermore, providing this data point would significantly simplify automated testing scenarios, enabling direct verification of task creation functionalities without resorting to workarounds. 3. **Modify the behavior** so adding a new entry preserves the user's current scroll position within the table. This would significantly improve the user experience and streamline task management. 4. **Clarify the intended purpose of the "Done Tasks" button** by implementing one of the expected behaviors below. This will provide users with clear functionality and avoid confusion. Clicking the "Done Tasks" button should trigger a specific action, such as: - Displaying a list of all tasks marked as done/completed by the user. - Taking users to a dedicated "Completed Tasks" section within the application. #################### File: TEST%20REPORT%20-%20Lois%20Layne%20Ramos.pdf Page: 11 Context: Additionally, consider providing a tooltip or hover text explaining the button's function on mouseover, further enhancing user experience. 5. The **"Add/Submit"** button should be disabled or visually grayed out when the input field is empty, clearly indicating that no action can be performed until data is entered. Optionally, a tooltip or error message could be displayed upon hovering over the disabled button, explaining the requirement for user input. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 2 Context: # A First Course in Complex Analysis Copyright 2002–2018 by the authors. All rights reserved. The most current version of this book is available at the website [http://math.sfsu.edu/beck/complex.html](http://math.sfsu.edu/beck/complex.html). This book may be freely reproduced and distributed, provided that it is reproduced in its entirety from the most recent version. This book may not be altered in any way, except for changes in format required for printing or other distribution, without the permission of the authors. This edition published with express permission of the authors by Orthogonal Publishing, Inc. Ann Arbor, Michigan [www.orthogonpublishing.com](http://www.orthogonpublishing.com) Typeset in Adobe Garamond Pro using Math Design mathematical fonts. ## About the cover The cover illustration, *Square Squared* by Robert Chaffin, shows two superimposed images. The foreground image represents the result of applying a transformation, \( z = z^2 \) (see Exercises 3.53 and 3.54), to the background image. The locally-conformable property of this mapping can be observed through matching the line segments, angles, and Sierpiński triangle features of the background image with their respective images in the foreground figure. (The foreground figure is scaled down to about 40% and repositioned to accommodate artistic and visibility considerations.) The background image fills the square with vertices at \( 0, 1, 1+i, \) and \( i \) (the positive direction along the imaginary axis is chosen as downward). It was prepared by using Michael Barnsley's chaos game, explaining on the fact that a square is self-similar, and by using a fractal–coloring method. A subset of the image is seen as a standard Sierpiński triangle. The chaos game was also re-purposed to create the foreground image. Robert Chaffin is Professor Emeritus at Central Michigan University. His academic interests are in abstract algebra, combinatorics, geometry, and computer applications. Some retirement from teaching has allowed much of this time to applying those interests to creation of art images. [http://people.cst.cmich.edu/chaffirn/Art_From_Mathematics/](http://people.cst.cmich.edu/chaffirn/Art_From_Mathematics/) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 3 Context: # "And what is the use of a book," thought Alice, "without pictures or conversations?" Lewis Carroll (Alice in Wonderland) ## About this book *A First Course in Complex Analysis* was written for a one-semester undergraduate course developed at Binghamton University (SUNY) and San Francisco State University, and has been adopted at several other institutions. For many of our students, Complex Analysis is their first rigorous analysis (if not mathematics) class they take, and this book reflects this very much. We tried to rely on as few concepts from real analysis as possible. In particular, series and sequences are treated from scratch, which has the consequence that power series are introduced late in the course. The goal our book works toward is the Residue Theorem, including some nontraditional applications from both continuous and discrete mathematics. A printed paperback version of this open textbook is available from Orthogonal Publishing (www.orthogonalpublishing.com) or your favorite online bookseller. ## About the authors **Matthias Beck** is a professor in the Mathematics Department at San Francisco State University. His research interests are in geometric combinatorics and analytic number theory. He is the author of three other books, *Computing the Continuous Discretion: Integer-point Enumeration in Polyhedra* (with Sinal Robins, Springer 2007), *The Art of Proof: Basic Training for Mathematics* (with Ross Geoghegan, Springer 2010), and *Combinatorial Reciprocity Theorem: An Invitation to Enumerative Geometric Combinatorics* (with Raman Sanyal, AMS 2018). **Gerald M. Green** is a lecturer in the Department of Mathematical Sciences at Binghamton University (SUNY). **Dennis P. Stinson** is a professor emeritus in the Department of Mathematical Sciences at Binghamton University (SUNY). His research interests are in dynamical systems and formal languages. **Lucas Sabalka** is an applied mathematician at a technology company in Lincoln, Nebraska. He works on 3-dimensional computer vision applications. He was formerly a professor of mathematics at St. Louis University, after postdoctoral positions at UC Davis and Binghamton University (SUNY). His mathematical research interests are in geometric group theory, low dimensional topology, and computational algebra. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 4 Context: # Robert Chaffee Robert Chaffee (cover art) is a professor emeritus at Central Michigan University. His academic interests are in abstract algebra, combinatorics, geometry, and computer applications. Since retirement from teaching, he has devoted much of his time to applying those interests to creation of art images. ## A Note to Instructors The material in this book should be more than enough for a typical semester-long undergraduate course in complex analysis; our experience teaches us that there is more content in this book than fits into one semester. Depending on the nature of your course and its place in your department's overall curriculum, some sections can be either partially omitted or their definitions and theorems can be assumed true without delving into proofs. Chapter 10 contains optional longer homework problems that could also be used as group projects at the end of a course. We would be happy to hear from anyone who has adopted our book for their course, as well as suggestions, corrections, or other comments. ## Acknowledgments We thank our students who made many suggestions for and found errors in the text. Special thanks go to Sheldon Axler, Collin Bleak, Pierre-Alexandre Bliman, Matthew Brin, Andrew Huang, John McCleary, Sharma Pallakonda, Joshua Palamater, and Dmytro Savchuk for comments, suggestions, and additions after teaching from this book. We thank Lon Mitchell for his initiative and support for the print version of our book with Orthogonal Publishing, and Bob Chaffee for allowing us to feature his art on the book’s cover. We are grateful to the American Institute of Mathematics for including our book in their Open Textbook Initiative [https://ainth.org/textbooks](https://ainth.org/textbooks). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 5 Context: # Contents 1. **Complex Numbers** 1.1. Definitions and Algebraic Properties . . . . . . . . 2 1.2. From Algebra to Geometry and Back . . . . . . . . 5 1.3. Geometric Properties . . . . . . . . . . . . . . . . 9 1.4. Elementary Topology of the Plane . . . . . . . . . 12 Optional Lab . . . . . . . . . . . . . . . . . . . . . . 21 2. **Differentiation** 2.1. Limits and Continuity . . . . . . . . . . . . . . . . 23 2.2. Differentiability and Holomorphicity . . . . . . . 28 2.3. The Cauchy–Riemann Equations . . . . . . . . . . . 32 2.4. Constant Functions . . . . . . . . . . . . . . . . . 36 3. **Examples of Functions** 3.1. Möbius Transformations . . . . . . . . . . . . . . . 43 3.2. Infinity and the Cross Ratio . . . . . . . . . . . 46 3.3. Stereographic Projection . . . . . . . . . . . . . . 48 3.4. Exponential and Trigonometric Functions . . . . . 55 3.5. Logarithms and Complex Exponentials . . . . . . . 59 4. **Integration** 4.1. Definition and Basic Properties . . . . . . . . . . 71 4.2. Antiderivatives . . . . . . . . . . . . . . . . . . 76 4.3. Cauchy’s Theorem . . . . . . . . . . . . . . . . . 80 4.4. Cauchy’s Integral Formula . . . . . . . . . . . . 85 5. **Consequences of Cauchy’s Theorem** 5.1. Variations of a Theme . . . . . . . . . . . . . . . 97 5.2. Antiderivatives Again . . . . . . . . . . . . . . 100 5.3. Taking Cauchy's Formulas to the Limit . . . . . . 102 #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 6 Context: ``` # Contents 6. Harmonic Functions ................................. 110 6.1 Definitions and Basic Properties ................. 114 6.2 Mean-Value Principle ............................ 114 7. Power Series ....................................... 121 7.1 Sequences and Completeness ...................... 122 7.2 Series .......................................... 125 7.3 Sequences and Series of Functions ............... 131 7.4 Regions of Convergence .......................... 135 8. Taylor and Laurent Series .......................... 146 8.1 Power Series and Holomorphic Functions .......... 146 8.2 Classification of Zeros and the Identity Principle 152 8.3 Laurent Series .................................. 156 9. Isolated Singularities and the Residue Theorem..... 169 9.1 Classification of Singularities ................... 169 9.2 Residues ........................................ 176 9.3 Argument Principle and Rouché’s Theorem ........ 180 10. Discrete Applications of the Residue Theorem ...... 188 10.1 Infinite Sums ................................... 188 10.2 Binomial Coefficients ........................... 189 10.3 Fibonacci Numbers ............................... 190 10.4 The Coin-Exchange Problem ...................... 193 10.5 Dedekind Sums .................................. 193 ## Theorems From Calculus ## Solutions to Selected Exercises ## Index ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 7 Context: # Chapter 1 ## Complex Numbers Die ganzen Zahlen hat der liebe Gott geschaffen, alles andere ist Menschenwerk. (God created the integers, everything else is made by humans.) Leopold Kronecker (1823–1891) The real numbers have many useful properties. There are operations such as addition, subtraction, and multiplication, as well as division by any nonzero number. There are useful laws that govern these operations, such as the commutative and distributive laws. We can take limits and do calculus, differentiating and integrating functions. But you cannot take a square root of \(-1\); that is, you cannot find a real root of the equation $$ x^2 + 1 = 0. \tag{1.1} $$ Most of you have heard that there is a "new" number that is a root of (1.1); that is, \(x^2 + 1 = 0\) or \(x^2 = -1\). We will show that when the real numbers are enlarged to a new system called the **complex numbers**, which includes \(i\), not only do we gain numbers with interesting properties, but we do not lose many of the nice properties that we had before. The complex numbers, like the real numbers, will have the operations of addition, subtraction, multiplication, as well as division by any complex number except zero. These operations will follow all the laws that we are used to, such as the commutative and distributive laws. We will also be able to take limits and do calculus. And there will be a root of (1.1). As a brief historical aside, complex numbers did not originate with the search for a square root of \(-1\); rather, they were introduced in the context of cubic equations. Scipione del Ferro (1465–1526) and Niccolò Tartaglia (1499–1557) discovered a way to find a root of any cubic polynomial, which was publicized by Gerolamo Cardano (1501–1576) and is often referred to as **Cardano's formula**. For the cubic polynomial \(x^3 + px + q\), Cardano's formula involves the quantity $$ \sqrt[3]{-\frac{q}{2} + \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}}. $$ #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 8 Context: ``` # Complex Numbers It is not hard to come up with examples for \( p \) and \( q \) for which the argument of this square root becomes negative and thus not computable within the real numbers. On the other hand (e.g., by arguing through the graph of a cubic polynomial), every cubic polynomial has at least one real root. This seeming contradiction can be solved using complex numbers, as was probably first exemplified by Rafael Bombelli (1526–1572). In the next section we show exactly how the complex numbers are set up, and in the rest of this chapter we will explore the properties of the complex numbers. These properties will be of both algebraic (such as the commutative and distributive properties mentioned already) and geometric nature. You will see, for example, that multiplication can be described geometrically. In the rest of the book, the calculus of complex numbers will be built on the properties that we develop in this chapter. ## 1.1 Definitions and Algebraic Properties There are many equivalent ways to think about a complex number, each of which is useful in its own right. In this section, we begin with a formal definition of a complex number. We then interpret this formal definition in more useful and easier-to-work-with algebraic language. Later we will see several more ways of thinking about complex numbers. **Definition.** The complex numbers are pairs of real numbers: \[ C = \{(x, y) : x, y \in \mathbb{R} \} \] equipped with the addition \[ (x, y) + (a, b) = (x + a, y + b) \quad (1.2) \] and the multiplication \[ (x, y) \cdot (a, b) = (x a - y b, x b + y a) \quad (1.3) \] One reason to believe that the definitions of these binary operations are acceptable is that \( C \) is an extension of \( \mathbb{R} \), in the sense that the complex numbers of the form \((x, 0)\) behave just like real numbers: \[ (x, 0) + (y, 0) = (x + y, 0) \quad \text{and} \quad (x, 0) \cdot (y, 0) = (xy, 0). \] ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 9 Context: So we can think of the real numbers being embedded in \( \mathbb{C} \) as those complex numbers whose second coordinate is zero. The following result states the algebraic structure that we established with our definitions. ## Proposition 1.1 Let \( (x, y) \) be a field, that is, for all \( (a, b) \in \mathbb{C} \): 1. \( (x, y) + (a, b) \in \mathbb{C} \) (1.4) 2. \( (x, y) + (a, d) + (b, c) = (x, y) + ((a + b), c) \) (1.5) 3. \( (x, y) + (0, 0) = (x, y) \) (1.6) 4. \( (x, y) + (-x, -y) = (0, 0) \) (1.7) 5. \( (a, b) + (c, d) = (x, y) \cdot (a, b) + (c, d) \) (1.8) 6. \( (x, y) \in \mathbb{C} \) (1.9) 7. \( (x, y) \cdot (a, b) = (x \cdot a, y \cdot b) \) (1.10) 8. \( (x, y) \cdot (b, d) = (x \cdot b, (a \cdot b)(c, d)) \) (1.11) 9. \( (x, y) \cdot (a, b) = (x, y) \) (1.12) 10. \( (x, y) \cdot (1, 0) = (x, y) \) (1.13) 11. For all \( (x, y) \in \mathbb{C} \setminus \{(0, 0)\} \): \( (x, y) \cdot \left( \frac{(y, -x)}{x^2 + y^2} \right) = (1, 0) \) (1.14) What we are stating here can be compressed in the language of algebra: equations (1.4)–(1.8) say that \( (\mathbb{C}, +) \) is an Abelian group with identity \( (0, 0) \); equations (1.10)–(1.14) say that \( (\mathbb{C} \setminus \{(0, 0)\}, \cdot) \) is an Abelian group with identity \( (1, 0) \). The proof of Proposition 1.1 is straightforward but nevertheless makes for good practice (Exercise 1.14). We give one sample: **Proof (1.18)**. By our definition for complex addition and properties of additive inverse in \( \mathbb{R} \): \[ (x, y) + (-x, -y) = (x + (-x), y + (-y)) = (0, 0). \] The definition of our multiplication implies the innocent-looking statement: \[ (0, 1) \cdot (1, 0) = (1, 0). \quad (1.15) \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 10 Context: ``` # Complex Numbers This identity together with the fact that $$ (x_0, \cdots, x_n) \cdot (x, y) = (ax, az) $$ allows an alternative notation for complex numbers. The latter implies that we can write $$ (x, y) = (x_0, 0) + (0, y) = (x_0, 0) + (0, 1)(0, y). $$ If we think—in the spirit of our remark about embedding R into C—of (x, 0) and (y, 0) as the real numbers x and y, then this means that we can write any complex number (x, y) as a linear combination of (1, 0) and (0, 1), with the real coefficients x and y. Now (1, 0), in turn, can be thought of as the real number 1. So if we give (0, 1) a special name, say i, then the complex number that we used to call (x, y) can be written as $$ x + iy. $$ ### Definition The number x is called the **real part** and y the **imaginary part** of the complex number x + iy, often denoted as Re(x + iy) = x and Im(x + iy) = y. The identity (1.15) then reads $$ i^2 = -1. $$ In fact, much more can now be said with the introduction of the square root of -1. It is not just that (1.1) has a root, but every nonconstant polynomial has roots in C: ### Fundamental Theorem of Algebra (see Theorem 5.11). Every nonconstant polynomial of degree d has d roots (counting multiplicity) in C. The proof of this theorem requires some (important) machinery, so we defer its proof and an extended discussion of it to Chapter 5. We invite you to check that the definitions of our binary operations and Proposition 1.1 are coherent with the usual real arithmetic rules if we think of complex numbers as given in the form x + iy. *The names have historical reasons: people thought of complex numbers as unreal, imagined.* ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 11 Context: # 1.2 From Algebra to Geometry and Back Although we just introduced a new way of writing complex numbers, let’s for a moment return to the \((x, y)\)-notation. It suggests that we can think of a complex number as a two-dimensional real vector. When plotting these vectors in the plane \(\mathbb{R}^2\), we will call the \(x\)-axis the real axis and the \(y\)-axis the imaginary axis. The addition that we defined for complex numbers resembles vector addition; see Figure 1.1. The analogy stops at multiplication: there is no “usual” multiplication of two vectors in \(\mathbb{R}^2\) that gives another vector, and certainly not one that agrees with our definition of the product of two complex numbers. ![Figure 1.1: Addition of complex numbers.](#) Any vector in \(\mathbb{R}^2\) is defined by its two coordinates. On the other hand, it is also determined by its length and the angle it encloses with, say, the positive real axis; let’s define these concepts thoroughly. ### Definition The absolute value (also called the modulus) of \(z = x + iy\) is \[ r = |z| = \sqrt{x^2 + y^2}, \] and an argument of \(z = x + iy\) is a number \(\phi \in \mathbb{R}\) such that \[ x = r \cos \phi \quad \text{and} \quad y = r \sin \phi. \] A given complex number \(z = x + iy\) has infinitely many possible arguments. For instance, the number \(1 = 1 + 0i\) lies on the positive real axis, and has argument \(0\), but we could just as well say it has argument \(2\pi, -2\pi, \text{or } 2k\pi\) for any integer \(k\). The number \(0 = 0 + 0i\) has modulus \(0\), and every real number \(y\) is an argument. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 12 Context: # COMPLEX NUMBERS Aside from the exceptional case of 0, for any complex number \( z \), the arguments of \( z \) all differ by a multiple of \( 2\pi \), just as we saw for the example \( z = 1 \). The absolute value of the difference of two vectors has a nice geometric interpretation: ## Proposition 1.2 Let \( z_1, z_2 \in \mathbb{C} \) be two complex numbers, thought of as vectors in \( \mathbb{R}^2 \), and let \( d(z_1, z_2) \) denote the distance between the endpoints of the two vectors in \( \mathbb{R}^2 \) (see Figure 1.2). Then \[ d(z_1, z_2) = |z_1 - z_2| = |z_2 - z_1|. \] **Proof:** Let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \). From geometry, we know that \[ d(z_1, z_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}. \] This is the definition of \( |z_1 - z_2| \). Since \( (x_1 - x_2) = (x_2 - x_1) \) and \( (y_1 - y_2) = (y_2 - y_1) \), this is also equal to \( |z_2 - z_1| \. \qed ![Figure 1.2: Geometry behind the distance between two complex numbers.](path_to_image) That \( |z_1 - z_2| = |z_2 - z_1| \) simply says that the vector from \( z_2 \) to \( z_1 \) has the same length as the vector from \( z_1 \) to \( z_2 \). One reason to introduce the absolute value and argument of a complex number is that they allow us to give a geometric interpretation for the multiplication of two complex numbers. Let’s say we have two complex numbers \( z_1 = x_1 + iy_1 \), with absolute value \( r_1 \) and argument \( \phi_1 \), and \( z_2 = x_2 + iy_2 \), with absolute value \( r_2 \) and argument \( \phi_2 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 13 Context: # FROM ALGEBRA TO GEOMETRY AND BACK ## 7 ϕ₂. This means we can write \(x_1 + iy_1 = (r_1 \cos \phi_1) + i(r_1 \sin \phi_1)\) and \(x_2 + iy_2 = (r_2 \cos \phi_2) + i(r_2 \sin \phi_2)\). To compute the product, we make use of some classic trigonometric identities: \[ (x_1 + iy_1)(x_2 + iy_2) = (r_1 \cos \phi_1 + ir_1 \sin \phi_1)(r_2 \cos \phi_2 + ir_2 \sin \phi_2) \] \[ = r_1 r_2 (\cos \phi_1 \cos \phi_2 - \sin \phi_1 \sin \phi_2) + i(r_1 r_2 \cos \phi_2 \sin \phi_1 + r_1 r_2 \sin \phi_2 \cos \phi_1) \] \[ = r_1 r_2 \left( \cos(\phi_1 + \phi_2) + i \sin(\phi_1 + \phi_2) \right). \] So the absolute value of the product is \(r_1 r_2\) and one of its arguments is \(ϕ_1 + ϕ_2\). Geometrically, we are multiplying the lengths of the two vectors representing our two complex numbers and adding their angles measured with respect to the positive real axis. ![Figure 1.3: Multiplication of complex numbers](#) In view of the above calculation, it should come as no surprise that we will have to deal with quantities of the form \(x + iy\) (where \(y\) is some real number) quite a bit. To save space, bytes, ink, etc., and because "Mathematics is for lazy people" we introduce a shortcut notation and define: \[ e^{i\phi} = \cos \phi + i \sin \phi. \] ^1 You should convince yourself that there is no problem with this fact since there are many possible arguments for complex numbers, as both cosine and sine are periodic functions with period \(2\pi\). ^2 Peter Halton (unpublished articles, Hudson River Undergraduate Mathematics Conference 2000). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 14 Context: # COMPLEX NUMBERS Figure 1.4 shows three examples. At this point, this exponential notation is indeed ![Figure 1.4: Three sample complex numbers of the form \( e^{x} \).](image-path) purely a notation. We will later see in Chapter 3 that it has an intimate connection to the complex exponential function. For now, we motivate this maybe strange seeming definition by collecting some of its properties: ## Proposition 1.3. For any \( x_1, x_2 \in \mathbb{R} \): 1. \( e^{(x_1 + x_2 i)} = e^{x_1} e^{x_2 i} \) 2. \( e^{0} = 1 \) 3. \( \frac{d}{dx} e^{x} = e^{x} \) 4. \( |e^{x}| = 1 \) 5. \( e^{i\pi} = -1 \) You are encouraged to prove them (Exercise 1.16); again, we give a sample. ## Proof of \( e^{x} \). By definition of \( e^{x} \): \[ \frac{d}{dx} e^{(x + i)} = -\sin{(x)} + \cos{(x)} = i ( \cos{(x)} + i\sin{(x)}) = i e^{(x+i)} \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 15 Context: # Geometric Properties Proposition 1.3 implies that \( z^{n} = 1 \) for any integers \( n \) and \( r > 0 \). Thus, numbers of the form \( z^{n} \) with \( z \in \mathbb{Q} \) play a pivotal role in solving equations of the form \( z^{n} = 1 \)—plenty of reason to give them a special name. **Definition.** A root of unity is of the form \( z^{n} \) for some integers \( n \) and \( r > 0 \). A root of unity \( z \) is a complex number \( z \) such that \( z^{n} = 1 \) for some positive integer \( n \). In this case, we call \( z \) an \( n \)th root of unity. If \( n \) is the smallest positive integer with the property \( z^{n} = 1 \), then \( z \) is a primitive \( n \)th root of unity. **Example 1.4.** The 4th roots of unity are \( 1 \) and \( i \) and \( e^{i\frac{\pi}{2}} \). The latter two are primitive 4th roots of unity. With our new notation, the sentence "the complex number \( x + iy \) has absolute value \( r \) and argument \( \theta \) now becomes the identity \[ x + iy = r e^{i\theta}. \] The left-hand side is often called the **rectangular form**, the right-hand side the **polar form** of this complex number. We now have five different ways of thinking about a complex number: the formal definition, in rectangular form, in polar form, and geometrically, using Cartesian coordinates or polar coordinates. Each of these five ways is useful in different situations, and translating between them is an essential ingredient in complex analysis. The five ways and their corresponding notation are listed in Figure 1.5. This list is not exhaustive; see, e.g., Exercise 1.21. ## 1.3 Geometric Properties From the chain of basic inequalities \[ -\sqrt{x^{2} + y^{2}} \leq -|z| \leq \sqrt{x^{2} + y^{2}} \leq |z| \leq \sqrt{x^{2} + y^{2}}, \] (or, alternatively, by arguing with basic geometric properties of triangles), we obtain the inequalities \[ -|z| \leq \text{Re}(z) \leq |z| \] and \[ -|z| \leq \text{Im}(z) \leq |z|. \tag{1.16} \] The square of the absolute value has the nice property \[ |x + iy|^{2} = x^{2} + y^{2} = (x + iy)(x - iy). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 16 Context: # COMPLEX NUMBERS | | Algebraic | Geometric | Exponential | |------------|------------------|------------------|---------------| | Formal | \(x + iy\) | Cartesian | \(re^{i\theta}\) | | | Rectangular | \(x\) | Polar | | | | \(y\) | | **Figure 1.5:** Five ways of thinking about a complex number. This is one of many reasons to give the process of passing from \(x + iy\) to \(x - iy\) a special name. ## Definition The number \(x - iy\) is the (complex) conjugate of \(x + iy\). We denote the conjugate by \[ x + iy = x - iy. \] Geometrically, conjugating \(z\) means reflecting the vector corresponding to \(z\) with respect to the real axis. The following collects some basic properties of the conjugate. ## Proposition 1.5 For any \(z_1, z_2, z_3 \in \mathbb{C}\): 1. \(z_1 \overline{z_2} = \overline{z_1 z_2}\) 2. \(\overline{z_1 z_2} = \overline{z_2} \overline{z_1}\) 3. \(\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_2}}{\overline{z_1}}\) 4. \(\overline{z} = z\) 5. \(|z| = |z|\) The proofs of these properties are easy (Exercise 1.22); once more we give a sample. ### Proof of (b) Let \(z_1 = x_1 + iy_1\) and \(z_2 = x_2 + iy_2\). Then \[ z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2) = (x_1x_2 - y_1y_2) + i(x_1y_2 + y_1x_2) = \overline{z_1} \overline{z_2}. \] Image Analysis: ### 1. Localization and Attribution - **Page**: This is a page from a document or book, numbered at the top left as "10." - **Image 1**: An illustrated table at the top of the page labeled as "Figure 1.5." - **Text Block**: Multiple paragraphs of text underneath the table, including definitions, propositions, and proofs. ### 2. Object Detection and Classification - **Image 1:** - **Objects**: A table with three rows and three columns. - **Categories**: Types of complex number representations. - **Key Features**: - **Row 1**: "Formal" corresponds to (x, y). - **Row 2**: "Algebraic" includes rectangular (x + iy) and exponential (re^iθ). - **Row 3**: "Geometric" includes cartesian and polar representations. ### 3. Scene and Activity Analysis - **Scene**: - The scene depicts a page from a textbook or academic document. - The key activity is educational content delivery, specifically discussing complex numbers. ### 4. Text Analysis - **Detected Text**: - Title: "COMPLEX NUMBERS." - Sections: - "Figure 1.5: Five ways of thinking..." - Definitions and propositions related to complex numbers. - Proof of a property. - **Analysis**: - The page discusses different representations of complex numbers, highlights their properties, and presents relevant mathematical proofs. ### 5. Diagram and Chart Analysis - **Image 1:** - **Content**: The table categorizes the representations of complex numbers. - **Axes/Scales**: Not applicable. - **Legends**: Explanatory text within the cells. - **Key Insights**: - Complex numbers can be represented in multiple forms: formal, algebraic, geometric. - The table is a clear and concise way to present these different perspectives. ### 6. Product Analysis - **Not applicable**: No products were depicted in the image. ### 7. Anomaly Detection - **Not applicable**: No anomalies in the visual content. ### 8. Color Analysis - **Dominant Colors**: Black text on a white background. - **Impact**: Standard colors for printed or digital educational content, aiding readability and focus. ### 9. Perspective and Composition - **Perspective**: Straight-on view, typical for document pages. - **Composition**: - The table (Image 1) is at the top, drawing initial attention. - Subsequent text follows a structured format, typical of academic documents. ### 10. Contextual Significance - **Contribution**: - The image and text provide fundamental knowledge about complex numbers. - It enriches the educational material by categorizing and defining complex number representations. ### 11. Metadata Analysis - **Not available**: No metadata provided from the image. ### 12. Graph and Trend Analysis - **Not applicable**: No graphs present. ### 13. Graph Numbers - **Not applicable**: No graphs present. ### Additional Aspects - **Process Flows and Descriptions**: - A defined process for understanding the conjugate of a complex number. - Steps to prove a proposition about complex numbers. - **Type Designations**: - Designates formal, algebraic, and geometric types of complex number representations. - **Trend and Interpretation**: - Emphasizes the versatility and various intuitions behind complex number representations. - **Tables**: - The table encapsulates the five types of complex number representations succinctly. This comprehensive examination focuses on the detailed aspects and provides a thorough understanding of the educational content on the described page. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 17 Context: # GEOMETRIC PROPERTIES We note that \( f(z) \) yields a neat formula for the inverse of a nonzero complex number, which is implicit already in (1.14): \[ z^{-1} = \frac{1}{|z|^2} \overline{z}. \] A famous geometric inequality (which holds, more generally, for vectors in \( \mathbb{R}^n \)) goes as follows. ## Proposition 1.6 (Triangle Inequality) For any \( z_1, z_2 \in \mathbb{C} \) we have \( |z_1 + z_2| \leq |z_1| + |z_2| \). By drawing a picture in the complex plane, you should be able to come up with a geometric proof of the triangle inequality. Here we proceed algebraically: ### Proof We make extensive use of Proposition 1.5: \[ |z_1 + z_2|^2 = |(z_1 + z_2)(\overline{z_1 + z_2})| = z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + z_1\overline{z_2} + z_2\overline{z_1} + |z_2|^2 \] \[ \leq |z_1|^2 + 2\Re(z_1\overline{z_2}) + |z_2|^2 \] \[ = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cdot \frac{\Re(z_1\overline{z_2})}{|z_1||z_2|} \leq (|z_1| + |z_2|)^2, \] where the inequality follows from (1.16). Taking square roots on the left- and right-hand sides proves our claim. For future reference we list several useful variants of the triangle inequality: ## Corollary 1.7 For \( z_1, z_2, \ldots, z_n \in \mathbb{C} \), we have the following relations: (a) The triangle inequality: \( |z_1 + z_2| \leq |z_1| + |z_2|. \) (b) The reverse triangle inequality: \( |z_1 + z_2| \geq ||z_1| - |z_2||. \) (c) The triangle inequality for sums: \[ \left| \sum_{k=1}^{n} z_k \right| \leq \sum_{k=1}^{n} |z_k|. \] Inequality (a) is just a rewrite of the original triangle inequality, using the fact that \( |z| = |z| \) and follows by induction. The proof of the reverse triangle inequality (b) is left as Exercise 1.25. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 18 Context: # 1.4 Elementary Topology of the Plane In Section 1.2 we saw that the complex numbers \( C \), which were initially defined algebraically, can be identified with the points in the Euclidean plane \( \mathbb{R}^2 \). In this section we collect some definitions and results concerning the topology of the plane. \[ \begin{array}{c} y \\ \text{C}(2+i,2) \\ \hspace{-30pt} \text{D}(-2, 1) \\ \hspace{80pt} 1 \\ \hspace{-30pt} x \end{array} \] **Figure 1.6:** Sample circle and disk. In Proposition 1.2, we interpreted \( |z - w| \) as the distance between the complex numbers \( z \) and \( w \), viewed as points in the plane. So if we fix a complex number \( a \) and a positive real number \( r \), then all \( z \in C \) satisfying \( |z - a| = r \) form the set of points at distance \( r \) from \( a \); this set is the circle with center \( a \) and radius \( r \), which we denote by \[ C(a, r) = \{ z \in C : |z - a| = r \} \] The inside of this circle is called the open disk with center \( a \) and radius \( r \); we use the notation \[ D(a, r) = \{ z \in C : |z - a| < r \} \] Note that \( D(a, r) \) does not include the points on \( C(a, r) \). Figure 1.6 illustrates these definitions. Next we need some terminology for talking about subsets of \( C \). **Definition.** Suppose \( G \) is a subset of \( C \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 19 Context: # ELEMENTARY TOPOLOGY OF THE PLANE ## 1. Types of Points (a) A point \( a \in G \) is an **interior point** of \( G \) if some open disk centered at \( a \) is a subset of \( G \). (b) A point \( b \in E \) is a **boundary point** of \( G \) if every open disk centered at \( b \) contains a point in \( G \) and also a point that is not in \( G \). (c) A point \( c \in C \) is an **accumulation point** of \( G \) if every open disk centered at \( c \) contains a point of \( G \) different from \( c \). (d) A point \( d \in G \) is an **isolated point** of \( G \) if some open disk centered at \( d \) contains no point of \( G \) other than \( d \). The idea is that if you don't move too far from an interior point of \( G \) then you remain in \( G \); but a boundary point you can make an arbitrarily small move and get to a point inside \( G \) and you can also make an arbitrarily small move and get to a point outside \( G \). ## Definition A set is **open** if all its points are interior points. A set is **closed** if it contains all its boundary points. ## Example 1.18 For \( r > 0 \) and \( a \in \mathbb{C} \), the sets \( \{ z \in \mathbb{C} : |z - a| < r \} \) and \( \{ z \in \mathbb{C} : |z - a| = r \} \) are open. The closed disk \[ \overline{D[a, r]} := \{ z \in \mathbb{C} : |z - a| \leq r \} \] is an example of a closed set. A given set might be neither open nor closed. The complex plane \( \mathbb{C} \) and the empty set \( \emptyset \) are the only sets that are both open and closed. ## Definition The **boundary** \( \partial G \) of a set \( G \) is the set of all boundary points of \( G \). The interior of \( G \) is the set of all interior points of \( G \). The closure of \( G \) is the set \( \overline{G} = G \cup \partial G \). ## Example 1.19 The closure of the open disk \( D[a, r] \) is \( \overline{D[a, r]} \). The boundary of \( D[a, r] \) is the circle \( \partial D[a, r] \). ## Definition The set \( G \) is **bounded** if \( G \subset D[0, R] \) for some \( R \). One notion that is somewhat subtle in the complex domain is the idea of **connectedness**. Intuitively, a set is connected if it is "in one piece." In a set \( R \), a set is connected if and only if it is an interval, so there is little reason to discuss the matter. However, in the plane there is a vast variety of connected subsets. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 20 Context: # Complex Numbers ![Figure 1.7: The intervals [0,1) and (1,2) are separated.](image_link) ## Definition Two sets \( X, Y \subset \mathbb{C} \) are separated if there are disjoint open sets \( A, B \subset C \) so that \( X \subset A \) and \( Y \subset B \). A set \( G \subset \mathbb{C} \) is connected if it is impossible to find two separated nonempty sets whose union is \( G \). A region is a connected open set. The idea of separation is that the two open sets \( A \) and \( B \) ensure that \( X \) and \( Y \) cannot just "stick together." It is usually easy to check that a set is not connected. On the other hand, it is hard to use the above definition to show that a set is connected, since we have to rule out any possible separation. ## Example 1.10 The intervals \( X = [0,1) \) and \( Y = (1,2) \) on the real axis are separated: There are infinitely many choices for \( A \) and \( B \) that work; one choice is \( A = D[0,1) \) and \( B = D(1,2) \), depicted in Figure 1.7. Hence \( X \cup Y = [0,1) \cup (1,2) \) is not connected. One type of connected set that we will use frequently is a path. ## Definition A path (or curve) in \( \mathbb{C} \) is a continuous function \( \gamma : [a,b] \to \mathbb{C} \), where \([a,b]\) is a closed interval in \( \mathbb{R} \). We may think of \( \gamma \) as a parametrization of the image that is painted by the path and will often write this parametrization as \( \gamma(t) \), for \( a \leq t \leq b \). The path is smooth if it is differentiable and the derivative \( \gamma' \) is continuous and nonzero. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 21 Context: # ELEMENTARY TOPOLOGY OF THE PLANE This definition uses the calculus notions of continuity and differentiability; that is, \( \gamma: [a, b] \to C \) being continuous means that for all \( t_0 \in [a, b] \) \[ \lim_{t \to t_0} \gamma(t) = \gamma(t_0) \] and the derivative of \( \gamma \) at \( t_0 \) is defined by \[ \gamma'(t_0) = \lim_{t \to t_0} \frac{\gamma(t) - \gamma(t_0)}{t - t_0} \] \[ \gamma_1(t) = -2 + 2e^{it}, \quad 0 \leq t \leq 2\pi \] \[ \gamma_2(t) = \begin{cases} 3 + i(t - 2) & \text{if } 0 \leq t \leq 3 \\ 6 - i + i(t - 1) & \text{if } 3 < t \leq 5 \end{cases} \] Figure 1.8: Two paths and their parameterizations: \( \gamma_1 \) is smooth and \( \gamma_2 \) is continuous and piecewise smooth. Figure 1.8 shows two examples. We remark that each path comes with an orientation, i.e., a sense of direction. For example, the path \( \gamma_1 \) in Figure 1.8 is different from \[ \gamma_1(t) = -2 + 2e^{it}, \quad 0 \leq t \leq \frac{3\pi}{2} \] even though both \( \gamma_1 \) and \( \gamma_3 \) yield the same picture; \( \gamma_1 \) features a counter-clockwise orientation, where as that of \( \gamma_2 \) is clockwise. It is a customary and practical abuse of notation to use the same letter for the path and its parameterization. We emphasize that a path must have a parameterization, and that the parameterization must be defined and continuous on a closed and bounded interval \([a, b]\). Since topologically we may identify \( C \) with \( \mathbb{R}^2 \), a path can be specified. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 22 Context: ``` # Complex Numbers by giving two continuous real-valued functions of a real variable, \( x(t) \) and \( y(t) \), and setting \( y(t) = x(t) + iy(t) \). ## Definition The path \( \gamma: [a, b] \to C \) is simple if \( \gamma(t_1) = \gamma(t_2) \) is one-to-one, with the possible exception that \( \gamma(a) = \gamma(b) \) in plain English: the path does not cross itself. A path \( \gamma: [a, b] \to C \) is closed if \( \gamma(a) = \gamma(b) \). ## Example 1.11 The unit circle \( C[0,1] \), parametrized, e.g., by \( \gamma(t) = e^{it} \), \( 0 \leq t < 2\pi \), is a simple closed path. As seems intuitively clear, any path is connected; however, a proof of this fact requires a bit more preparation in topology. The same goes for the following result, which gives a useful property of open connected sets. ## Theorem 1.12 If any two points in \( G \subset C \) can be connected by a path in \( G \), then \( G \) is connected. Conversely, if \( G \subset C \) is open and connected, then any two points of \( G \) can be connected by a path in \( G \); in fact, we can connect any two points of \( G \) by a chain of horizontal and vertical segments lying in \( G \). Here a chain of segments in \( G \) means the following: there are points \( z_{0}, z_{1}, \ldots, z_{n} \) such that \( z_{0} \) and \( z_{n} \) are the endpoints of a horizontal or vertical segment in \( G \) for all \( k = 0, 1, \ldots, n-1 \) (it is not hard to parametrize such a chain, so it determines a path). ## Example 1.13 Consider the open unit disk \( D[0, 1) \). Any two points in \( D[0, 1) \) can be connected by a chain of at most two segments in \( D[0, 1) \) and so \( D[0, 1) \) is connected. Now let \( G = (D[0, 1) \setminus \{0\}) \). This is the punctured disk obtained by removing the center from \( D[0, 1) \). Then \( G \) is open and it is connected, but now you may need more than two segments to connect points. For example, you need three segments to connect \( z_{0} \to z_{1} \to z_{2} \) since we cannot go through \( 0 \). We remark that the second part of Theorem 1.12 is not generally true if \( G \) is not open. For example, circles are connected but there is no way to connect two distinct points of a circle by a chain of segments that are subsets of the circle. A more extreme example, discussed in topology texts, is the "topologist's sine curve," which is a connected set \( S \) that contains points that cannot be connected by a path of any sort within \( S \). ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 23 Context: # Exercises ## 1.1. Let \( z = 1 + 2i \) and \( w = 2 - i \). Compute the following: (a) \( z + 3w \) (b) \( w - z \) (c) \( z^2 \) (d) \( \text{Re}(wz + w) \) (e) \( z^2 + z + i \) ## 1.2. Find the real and imaginary parts of each of the following: (a) \( \frac{e^{ia}}{1 + i} \) for any \( a \in \mathbb{R} \) (b) \( 1 - i \) (c) \( \left( \frac{1}{\sqrt{2}} \right)^3 \) (d) \( i^n \) for any \( n \in \mathbb{Z} \) ## 1.3. Find the absolute value and conjugate of each of the following: (a) \( -2 + i \) (b) \( (2 + i\sqrt{3}) \) (c) \( \frac{3}{\sqrt{5}} - \sqrt{5}i \) (d) \( (1 + i)^{6} \) ## 1.4. Write in polar form: (a) \( 2i \) (b) \( 1 + i \) (c) \( 3 + \sqrt{3}i \) (d) \( -i \) (e) \( (2 - i)^2 \) (f) \( [3 - 4i] \) (g) \( \sqrt{5} - i \) (h) \( \left( \frac{1}{4} \right)^{4} \) ## 1.5. Write in rectangular form: (a) \( \sqrt{2}e^{i\frac{\pi}{4}} \) (b) \( 3e^{i\frac{\pi}{6}} \) (c) \( -r^{2}e^{i2\theta} \) (d) \( 2e^{i\frac{\pi}{3}} \) ## 1.6. Write in both polar and rectangular form: (a) \( e^{i\frac{\pi}{6}} \) (b) \( \frac{1}{2} e^{i\theta} \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 24 Context: # COMPLEX NUMBERS ## 1.7 Show that the quadratic formula works. That is, for \( a, b, c \in \mathbb{R} \) with \( a \neq 0 \), prove that the roots of the equation \( ax^2 + bx + c = 0 \) are \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here we define \( \sqrt{b^2 - 4ac} = i \sqrt{-(b^2 - 4ac)} \) if the discriminant \( b^2 - 4ac \) is negative. ## 1.8 Use the quadratic formula to solve the following equations: - (a) \( x^2 + 25 = 0 \) - (b) \( 2x^2 + 2x + 5 = 0 \) - (c) \( 5x^2 + 4x + 1 = 0 \) - (d) \( x^2 - 2x = 0 \) - (e) \( x^2 = 2x \) ## 1.9 Find all solutions of the equation \( x^2 + 2x + (1-i) = 0 \). ## 1.10 Fix \( a \in \mathbb{C} \) and \( b \in \mathbb{R} \). Show that the equation \( |z|^2 + \text{Re}(z) + b = 0 \) has a solution if and only if \( |a|^2 \geq 4b \). When solutions exist, show the solution set is a circle. ## 1.11 Find all solutions to the following equations: - (a) \( z^6 = 1 \) - (b) \( z^6 = -16 \) - (c) \( z^6 = e^{-3} \) - (d) \( z^6 - z^2 = 0 \) ## 1.12 Show that \( |z| = 1 \) if and only if \( z = \bar{z} \). ## 1.13 Show that: - (a) \( z \) is a real number if and only if \( z = \bar{z} \). - (b) \( z \) is either real or purely imaginary if and only if \( |z|^2 = z^2 \). ## 1.14 Review Proposition 1.1. ## 1.15 Show that if \( z_1 z_2 = 0 \) then \( z_1 = 0 \) or \( z_2 = 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 25 Context: # Elementary Topology of the Plane ## 1.16 Prove Proposition 1.3. ## 1.17 Fix a positive integer \( n \). Prove that the solutions to the equation \( z^n = 1 \) are precisely \( z = e^{2 \pi i k/n} \) where \( k \in \mathbb{Z} \). (Hint: To show that every solution of \( z^n = 1 \) is of this form, first prove that it must be of the form \( z = e^{i \theta} \) for some \( \theta \in \mathbb{R} \), then write \( z = m + ib \) for some integer \( m \) and some real number \( 0 < b < 1 \), and then argue that \( b \) has to be zero.) ## 1.18 Show that \[ z^2 - 1 = (z - 1)\left(z^2 + 2 \cos \frac{\pi}{3}(z^2 - 2z \cos \frac{\pi}{3} + 1\right) \] and deduce from this closed formula for \( \cos \frac{\pi}{3} \) and \( \cos \frac{2\pi}{3} \). ## 1.19 Fix a positive integer \( n \) and a complex number \( w \). Find all solutions to \( z^n = w \). (Hint: Write \( w \) in terms of polar coordinates.) ## 1.20 Use Proposition 1.3 to derive the triple angle formulas: (a) \(\cos(3\phi) = 4\cos^3\phi - 3\cos\phi \sin^2\phi\) (b) \(\sin(3\phi) = 3\cos^2\phi \sin\phi - \sin^3\phi\) ## 1.21 Given \( x, y \in \mathbb{R} \), define the matrix \( M(x, y) = \begin{bmatrix} x & -y \\ y & x \end{bmatrix} \). Show that \[ M(x, y) + M(a, b) = M(x + a, y + b) \] and \[ M(x, y)M(a, b) = M(xa - yb, xb + ya) \] (This means that the set \( \{ M(x, y) : x, y \in \mathbb{R} \} \) equipped with the usual addition and multiplication of matrices, behaves exactly like \( C = \{ (x, y) : x, y \in \mathbb{R} \} \).) ## 1.22 Prove Proposition 1.5. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 26 Context: # COMPLEX NUMBERS ## 1.23 Sketch the following sets in the complex plane: - (a) \( \{ z \in \mathbb{C} : |z - 1 + i| = 1 \} \) - (b) \( \{ z \in \mathbb{C} : |z - 1 + i| \leq 2 \} \) - (c) \( \{ z \in \mathbb{C} : \text{Re}(z^2 + 2 - 2i) = 3 \} \) - (d) \( \{ z \in \mathbb{C} : |z - i| + |z + i| = 3 \} \) - (e) \( \{ z \in \mathbb{C} : |z| = |z - 1| \} \) - (f) \( \{ z \in \mathbb{C} : |z - 1| = 2 |z + 1| \} \) - (g) \( \{ z \in \mathbb{C} : \text{Re}(z) = 1 \} \) - (h) \( \{ z \in \mathbb{C} : \text{Im}(z) = 1 \} \) ## 1.24 Suppose \( \rho \) is a polynomial with real coefficients. Prove that: - (a) \( \rho(\overline{z}) = \overline{\rho(z)} \). - (b) \( \rho(z) = 0 \) if and only if \( \rho(\overline{z}) = 0 \). ## 1.25 Prove the reverse triangle inequality (Proposition 1.7(b)): \[ |z_1 - z_2| \leq |z_1| + |z_2|. \] ## 1.26 Use the previous exercise to show that \[ \left| \frac{1}{z - 2} \right| < \frac{1}{3} \] for every \( z \) on the circle \( C[0,2] \). ## 1.27 Sketch the sets defined by the following constraints and determine whether they are open, closed, or neither; bounded; connected. - (a) \( |z + 3| < 2 \) - (b) \( \text{Im}(z) < 1 \) - (c) \( 0 < |z - 1| < 2 \) - (d) \( |z| = |1 + i| + 2 \) - (e) \( |z| = |1 + i| + 3 \) ## 1.28 What are the boundaries of the sets in the previous exercise? ## 1.29 Let \( G \) be the set of points \( z \in \mathbb{C} \) satisfying either \( z = r \) and \( -2 < r < -1 \), or \( |z| < 1 \), or \( z = \sigma \) where \( \sigma = -2 \). - (a) Sketch the set \( G \), being careful to indicate exactly the points that are in \( G \). - (b) Determine the interior points of \( G \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 27 Context: # Elementary Topology of the Plane 21 (c) Determine the boundary points of \( G \). (d) Determine the isolated points of \( G \). ## 1.30 The set \( G \) in the previous exercise can be written in three different ways as the union of two disjoint nonempty separated subsets. Describe them, and in each case say briefly why the subsets are separated. ## 1.31 Show that the union of two regions with nonempty intersection is itself a region. ## 1.32 Show that if \( A \subset B \) and \( B \) is closed, then \( A \subset B \). Similarly, if \( A \subset B \) and \( A \) is open, show that \( A \) is contained in the interior of \( B \). ## 1.33 Find a parameterization for each of the following paths: (a) The circle \( C(1 + i, 1) \), oriented counter-clockwise (b) The line segment from \( -1 - i \) to \( 2i \) (c) The top half of the circle \( C(0, \frac{3}{4}) \), oriented clockwise (d) The rectangle with vertices \( \pm 1 \pm 2i \), oriented counter-clockwise (e) The ellipse \( \{ z \in \mathbb{C} : |z - 1| + |z + 1| = 4 \} \), oriented counter-clockwise ## 1.34 Draw the path parameterized by \[ \gamma(t) = \cos(t) \cdot \cosh(t) + i \cdot \sin(t) \cdot \sinh(t), \quad 0 \leq t \leq 2\pi. \] ## 1.35 Let \( G \) be the annulus determined by the inequalities \( 2 < |z| < 3 \). This is a connected open set. Find the maximum number of horizontal and vertical segments in \( G \) needed to connect two points of \( G \). --- ### Optional Lab Open your favorite web browser and search for the complex function grapher for the open-source software GeoGebra. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 28 Context: # Complex Numbers 1. Convert the following complex numbers into their polar representation, i.e., give the absolute value and the argument of the number: \[ 3 + 4i = r e^{i\theta} = 2 + 2i = \frac{1}{\sqrt{3 + i}} = \] After you have finished computing these numbers, check your answers with the program. 2. Convert the following complex numbers given in polar representation into their rectangular representation: \[ 2e^{i\theta} = 3e^{i\phi} = \frac{1}{e^{i\varphi}} = -\sqrt{2}e^{2i\phi} = \] After you have finished computing these numbers, check your answers with the program. 3. Pick your favorite five numbers from the ones that you’ve played around with and put them in the tables below, in both rectangular and polar form. Apply the functions listed to your numbers. Think about which representation is more helpful in each instance. | rectangular | polar | |-------------|------------| | z + 1 | | | z + 2 - i | | | 2z | | | -z | | | z^2 | | | Re(z) | | | Im(z) | | | /Im(z) | | | |z| | | | 1/2 | | 4. Play with other examples until you get a feel for these functions. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 29 Context: # Chapter 2 ## Differentiation Mathematical study and research are very suggestive of generalizing. Whispers make several efforts before he climbed the Mathematics in the 1860s and even then it cut the life short of this party. Now, however, any source can be looked up for a small act, and perhaps one does not appreciate the difficulty of the original account. So in mathematics, it may be found hard to realize the great initial difficulty of making a little step which may seem so natural and obvious, and it may not be surprising if such a step has been found and lost again. **Louis Nodel (1888–1972)** We will now start our study of complex functions. The fundamental concept on which all of calculus is based is that of a limit—it allows us to develop the central properties of continuity and differentiability of functions. Our goal in this chapter is to do the same for complex functions. ### 2.1 Limits and Continuity **Definition.** A (complex) function \( f \) is a map from a subset \( G \subset \mathbb{C} \) to \( \mathbb{C} \); in this situation we will write \( f : G \to \mathbb{C} \) and call \( G \) the domain of \( f \). This means that each element \( z \in G \) gets mapped to exactly one complex number, called the image of \( z \) and usually denoted by \( f(z) \). So far there is nothing that makes complex functions any more special than, say, functions from \( \mathbb{R} \) to \( \mathbb{R} \). In fact, we can construct many familiar looking functions from the standard calculus repertoire, such as \( f(z) = z \) (the identity map), \( f(z) = z^2 + 1 \), or \( f(z) = \frac{1}{z} \). The former three could be defined on all of \( \mathbb{C} \), whereas for the latter we have to exclude the origin \( z = 0 \) from the domain. On the other hand, we could construct some functions that make use of a certain representation of \( z \), for example, \( f(x, y) = x - 2iy \), \( f(x, y) = y - i \), or \( f(z, \varphi) = 2r e^{i\varphi} \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 30 Context: Next we define limits of a function. The philosophy of the following definition is not restricted to complex functions, but for sake of simplicity we limit it only for those functions. ## Definition Suppose \( f : G \to \mathbb{C} \) and \( z_0 \) is an accumulation point of \( G \). If \( \epsilon \) is a complex number such that for every \( \delta > 0 \) we can find \( \delta > 0 \) such that \( 0 < |z - z_0| < \delta \), then \[ \lim_{z \to z_0} f(z) = w_0. \] This definition is the same as is found in most calculus texts. The reason we require that \( z_0 \) is an accumulation point of the domain is just that we need to be sure that there are points \( z \) of the domain that are arbitrarily close to \( z_0 \). Just as in the real case, our definition (i.e., the part that says \( 0 < |z - z_0| \)) does not require that \( z_0 \) is in the domain of \( f \); the definition explicitly ignores the value of \( f(z_0) \). ## Example 2.1 Let's prove that \( \lim_{z \to 2} z^2 = 4 \). Given \( \epsilon > 0 \), we need to determine \( \delta > 0 \) such that \( 0 < |z - 2| < \delta \) implies \( |z^2 - 4| < \epsilon \). We rewrite: \[ |z^2 - 4| = |z - 2||z + 2|. \] If we choose \( \delta < 1 \), then the factor \( |z + 2| \) on the right can be bounded by 3 (draw a picture). This means that any \( \delta < \min\{1, \frac{\epsilon}{3}\} \) should do the trick; in this case, \( 0 < |z - 2| < \delta \) implies: \[ |z^2 - 4| < 3 |z - 2| < 3 \delta < \epsilon. \] This proves \( \lim_{z \to 2} z^2 = 4 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 31 Context: # LIMITS AND CONTINUITY Just as in the real case, the limit \( w \) is unique if it exists (Exercise 2.3). It is often useful to investigate limits by restricting the way the point \( z \) approaches \( w \). The following result is a direct consequence of the definition. ## Proposition 2.2 Suppose \( f: G \rightarrow \mathbb{C} \) and \( \lim_{z \to w} f(z) = w_0 \), where \( w_0 \) is an accumulation point of \( G \). If \( f \) is the restriction of \( g \) to \( G \) then \( \lim_{z \to w} f(z) \) exists and has the value \( w_0 \). The definition of limit in the complex domain has to be treated with a little more care than its real companion; this is illustrated by the following example. ## Example 2.3 The limit of \( z^2 \) as \( z \to 0 \) does not exist. To see this, we try to compute this limit as \( z \to 0 \) on the real and on the imaginary axis. In the first case, we can write \( z = x \) where \( x \in \mathbb{R} \), and then \[ \lim_{x \to 0} \frac{1}{z} = \lim_{x \to 0} \frac{1}{x^2} = \lim_{x \to 0} \frac{1}{x} = 1. \] In the second case, we write \( z = iy \) where \( y \in \mathbb{R} \), and then \[ \lim_{y \to 0} \frac{1}{iy} = \lim_{y \to 0} \frac{-i}{y} = -1. \] So we get a different "limit" depending on the direction from which we approach \( 0 \). Proposition 2.2 then implies that the limit of \( z^2 \) as \( z \to 0 \) does not exist. On the other hand, the following usual limit rules are valid for complex functions; the proofs of these rules are everything but trivial and make for nice practice (Exercise 2.4); as usual, we give a sample proof. ## Proposition 2.4 Let \( f \) and \( g \) be complex functions with domain \( G \), let \( w \) be an accumulation point of \( G \), and let \( c \in \mathbb{C} \). If \( \lim_{z \to w} f(z) \) and \( \lim_{z \to w} g(z) \) exist, then 1. \( \lim_{z \to w} (f(z) + g(z)) = \lim_{z \to w} f(z) + \lim_{z \to w} g(z) \) 2. \( \lim_{z \to w} (f(z) g(z)) = \lim_{z \to w} f(z) \cdot \lim_{z \to w} g(z) \) 3. \( \lim_{z \to w} \left( \frac{f(z)}{g(z)} \right) = \frac{\lim_{z \to w} f(z)}{\lim_{z \to w} g(z)} \) where in the last identity we also require that \( \lim_{z \to w} g(z) \neq 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 32 Context: ``` # Differentiation Proof of **Theorem**. Assume that \( x_0 \in G \) (otherwise there is nothing to prove), and let \( L = \lim_{x \to x_0} f(x) \) and \( M = \lim_{x \to x_0} g(x) \). Then we know that given \( \epsilon > 0 \), we can find \( \delta_1, \delta_2 > 0 \) such that \[ 0 < |x - x_0| < \delta_1 \implies |f(x) - L| < \frac{\epsilon}{2} \] and \[ 0 < |x - x_0| < \delta_2 \implies |g(x) - M| < \frac{\epsilon}{2}. \] Thus, choosing \( \delta = \min(\delta_1, \delta_2) \), we infer that \( 0 < |x - x_0| < \delta \) implies \[ |f(x) + g(x) - (L + M)| \leq |f(x) - L| + |g(x) - M| < \epsilon. \] Here we used the triangle inequality (Proposition 1.6). This proves that \[ \lim_{x \to x_0} (f(x) + g(x)) = L + M, \] which was our claim. Because the definition of the limit is somewhat elaborate, the following fundamental definition looks almost trivial. ## Definition Suppose \( f : G \to \mathbb{C} \). If \( x_0 \in G \) and either \( x_0 \) is an isolated point of \( G \) or \[ \lim_{x \to x_0} f(x) = f(x_0) \] then \( f \) is continuous at \( x_0 \). More generally, \( f \) is continuous on \( E \subset G \) if it is continuous at every \( x \in E \). However, in almost all proofs using continuity it is necessary to interpret this in terms of \( \epsilon \) and \( \delta \). ## Definition Suppose \( f : G \to \mathbb{C} \) and \( x_0 \in G \). Then \( f \) is continuous at \( x_0 \) if, for every positive real number \( \epsilon \) there is a positive real number \( \delta > 0 \) such that \[ |f(x) - f(x_0)| < \epsilon \quad \text{for all } x \in E \text{ satisfying } |x - x_0| < \delta. \] See Exercise 2.11 for a proof that these definitions are equivalent. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 33 Context: # LIMITS AND CONTINUITY ## Example 2.5 We already proved (in Example 2.1) that the function \( f : C \to C \) given by \( f(x) = x^2 \) is continuous at \( x = 1 \). You're invited (Exercise 2.8) to extend our proof to show that, in fact, this function is continuous on \( C \). On the other hand, let \( g : C \to C \) be given by \[ g(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \] In Example 2.3 we proved that \( g \) is not continuous at \( x = 0 \). However, this is its only point of discontinuity (Exercise 2.9). Just as in the real case, we can "take the limit inside" a continuous function, by considering composition of functions. ## Definition The image of the function \( g : G \to C \) is the set \( \{ g(x) : x \in G \} \). If the image of \( g \) is contained in the domain of another function \( f : H \to C \), we define the composition \( f \circ g : G \to C \) through \[ (f \circ g)(x) = f(g(x)). \] ## Proposition 2.6 Let \( g : G \to C \) with image contained in \( H \), and let \( f : H \to C \). Suppose \( x_n \) is an accumulation point of \( G \), \( \lim_{n \to \infty} g(x_n) = w_0 \in H \), and \( f \) is continuous at \( w_0 \). Then \[ \lim_{n \to \infty} f(g(x_n)) = f\left(\lim_{n \to \infty} g(x_n)\right), \] in short, \[ \lim_{n \to \infty} f(g(x_n)) = f\left( \lim_{n \to \infty} g(x_n) \right). \] **Proof.** Given \( \epsilon > 0 \), we know there is \( \delta > 0 \) such that \[ |w - w_0| < \delta \implies |f(w) - f(w_0)| < \epsilon. \] For this, we also know there is a \( \eta > 0 \) such that \[ 0 < |x - z_0| < \eta \implies |g(x) - w| < \delta. \] Stringing these two implications together gives that \[ 0 < |x - z_0| < \eta \implies |f(g(x)) - f(w_0)| < \epsilon. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 34 Context: We have thus proved that \(\lim_{z \to z_0} f(z) = f(z_0)\). ## 2.2 Differentiability and Holomorphy The fact that simple functions such as \(\frac{1}{z}\) do not have limits at certain points illustrates something special about complex numbers that has no parallel in the reals—we can express a function in a very compact way in one variable, yet it shows some peculiar behavior in the limit. We will repeatedly notice this kind of behavior; one reason is that when trying to compute a limit of a function \(f(z)\), say, as \(z \to 0\), we have to allow to approach the point in any way. On the real line there are only two directions to approach \(0\)—from the left or from the right (or some combination of those two). In the complex plane, we have an additional dimension to play with. This means that the statement > A complex function has a limit... is in many senses stronger than the statement > A real function has a limit... This difference becomes apparent most badly when studying derivatives. **Definition.** Suppose \(f: G \to \mathbb{C}\) is a complex function and \(z_0\) is an interior point of \(G\). The derivative of \(f\) at \(z_0\) is defined as \[ f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \tag{2.1} \] provided this limit exists. In this case, \(f\) is called differentiable at \(z_0\). If \(f\) is differentiable for all points in an open disk centered at \(z_0\), then \(f\) is called holomorphic* at \(z_0\). The function \(f\) is holomorphic on the open set \(G \subset \mathbb{C}\) if it is differentiable (and hence holomorphic) at every point in \(E\). Functions that are differentiable (and hence holomorphic) in the whole complex plane \(C\) are called entire. **Example 2.7.** The function \(f: \mathbb{C} \to \mathbb{C}\) given by \(f(z) = z^2\) is entire, that is, holomorphic in \(C\) for any \(z \in \mathbb{C}\), \[ f'(z) = \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \lim_{h \to 0} \frac{(z+h)^2 - z^2}{h} = \lim_{h \to 0} \frac{2zh + h^2}{h} = 2z. \] *Some experts use the term analytic instead of holomorphic. As we will explain in Chapter 8, in our context, these two terms are synonymous. Technically, though, these terms have different definitions. Here we will be using the above definition; we will stick with using the term holomorphic instead of the term analytic. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 35 Context: # Differentiability and Holomorphicity The difference quotient limit (2.1) can be rewritten as $$ f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h} $$ This equivalent definition is sometimes easier to handle. Note that \( h \) need not be a real number but can rather approach zero from anywhere in the complex plane. The notions of differentiability and holomorphicity are not interchangeable: ## Example 2.8 The function \( f: \mathbb{C} \to \mathbb{C} \) given by \( f(z) = z^2 \) is differentiable at 0 and nowhere else; in particular, \( f \) is not holomorphic at 0. Let’s write \( z = r e^{i\phi} \). Then, \[ \frac{z^2 - z_0^2}{z - z_0} = \frac{(r e^{i\phi} + r e^{i\phi}) - z_0^2}{r e^{i\phi} - z_0} = \frac{2r^2 e^{2i\phi} - z_0^2}{r e^{i\phi} - z_0} \] When \( z_0 \neq 0 \), then taking the limit of \( f(z) \) as \( z \to z_0 \) thus means taking the limit of \[ \lim_{r \to 0} \frac{2z_0 r e^{i\phi} + r^2 e^{2i\phi} - z_0^2}{r e^{i\phi} - z_0} \] a number that depends on \( \phi \), i.e., on the direction that \( z \) approaches \( z_0 \). Hence this limit does not exist. On the other hand, if \( z_0 = 0 \) then the right-hand side above equals \( r^{1- \phi} = |z|^{1-\phi} \). Hence \[ \lim_{r \to 0} \left| \frac{z^2}{|z|^{2-\phi}} \right| = \lim_{r \to 0} \lim_{z \to 0} \frac{z^2}{0} = 0. \] Example 2.9 The function \( f: \mathbb{C} \to \mathbb{C} \) given by \( f(z) = \frac{z^2 - z_0^2}{z - z_0} \) is nowhere differentiable: \[ \lim_{z \to z_0} \frac{z^2 - z_0^2}{z - z_0} = \text{ does not exist, as discussed in Example 2.3. } #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 36 Context: # Differentiation The basic properties for derivatives are similar to those we know from real calculus. In fact, the following rules follow mostly from properties of the limit. ## Proposition 2.10 Suppose \( f \) and \( g \) are differentiable at \( c \in \mathbb{C} \) and \( g \) is differentiable at \( g(c) \). Then: 1. \( (f + g)'(z) = f'(z) + g'(z) \) for any \( z \in \mathbb{C} \) 2. \( (f \cdot g)'(z) = f'(z)g(z) + f(z)g'(z) \) 3. \( \left( \frac{f}{g} \right)'(z) = \frac{f'(z)g(z) - f(z)g'(z)}{(g(z))^2} \) provided that \( g(z) \neq 0 \) 4. \( (z^n)' = n z^{n-1} \) for any nonzero integer \( n \) 5. \( g \) is continuous at \( c \) 6. \( (h \circ g)'(z) = h'(g(z))g'(z) \) **Proof of (2):** \[ (fg)'(z) = \lim_{h \to 0} \frac{f(z + h)g(z + h) - f(z)g(z)}{h} \] \[ = \lim_{h \to 0} \left( f(z + h)g(z + h) - f(z)g(z + h) + f(z)g(z + h) - f(z)g(z) \right) \frac{1}{h} \] \[ = \lim_{h \to 0} \left( \frac{f(z + h) - f(z)}{h}g(z + h) + f(z)\frac{g(z + h) - g(z)}{h} \right) \] \[ = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} \cdot g(z) + f(z) \cdot \lim_{h \to 0} \frac{g(z + h) - g(z)}{h} \] \[ = f'(z)g(z) + f(z)g'(z) \] Note that we have used the definition of the derivative and Proposition 2.4(a) & (b) (the addition and multiplication rules for limits). A prominent application of the differentiation rules is the composition of a complex function \( f(z) \) with a path \( \gamma(t) \). The proof of the following result gives the reverse. ## Proposition 2.11 Suppose \( f \) is holomorphic at \( c \in \mathbb{C} \) with \( f(\gamma(0)) \) and \( \gamma_1 \) and \( \gamma_2 \) are two smooth paths that pass through \( c \), making an angle of \(\theta\) with each other. Then if \( \gamma_1 \) transforms \( z \) and \( \gamma_2 \) into smooth paths which meet \( f(\gamma) \), the transformed paths make an angle of \( \alpha \) with each other. In words, a holomorphic function with nonzero derivative preserves angles. Functions that preserve angles in this way are called conformal. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 37 Context: # Differentiability and Holomorphicity ## Proof Let \( \gamma_1(t) \) and \( \gamma_2(t) \) be parametrizations of the two paths such that \( \gamma_1(0) = \gamma_2(0) = \gamma(0) \). Then \( \gamma_2(t) \) (considered as a vector) is the tangent vector of \( \gamma_1 \) at the point \( a \), and \( \gamma_2'(0) \) is the tangent vector of \( \gamma_2 \) at the point \( f(a) \) given by: \[ \frac{d}{dt} \bigg|_{t=0} \gamma_1(t) = f'(\gamma_1(0)) \cdot \gamma_1'(0) \] and similarly, the tangent vector of \( f(\gamma_2) \) at the point \( f(a) \) is \( f'(\gamma_2'(0)) = f'[\gamma_2'(0)] \). This means that the action of \( f \) multiplies the two tangent vectors \( \gamma_1'(0) \) and \( \gamma_2'(0) \) by the same nonzero complex number \( f'(a) \), and so the two tangent vectors get dilated by \( |f'(a)| \) (which does not affect their direction) and rotated by the same angle (an argument of \( f(a) \)). We end this section with yet another differentiation rule, that for inverse functions. As in the real case, this rule is only defined for functions that are bijections. ## Definition A function \( f: G \to H \) is one-to-one if for every image \( w \in H \) there is a unique \( z \in G \) such that \( f(z) = w \). The function is onto if every \( w \in H \) has a preimage \( z \in G \) that gives \( f(z) = w \). A bijection is a function that is both one-to-one and onto. If \( f: G \to H \) is a bijection then \( g: H \to G \) is the inverse of \( f \) if \( f(g(z)) = z \) for all \( z \in H \); in other words, the composition \( g \circ f \) is the identity function on \( H \). ## Proposition 12 Suppose \( G, H \subset \mathbb{C} \) are open sets, \( f: G \to H \) is a bijection, \( s: H \to G \) is the inverse function of \( f \), and \( z_0 \in H \). If \( f \) is differentiable at \( s(z_0) \) with \( f'(s(z_0)) \neq 0 \) and \( g \) is continuous at \( z_0 \), then \( g \) is differentiable at \( z_0 \), with \[ g'(z_0) = \frac{1}{f'(s(z_0))} \] ## Proof Since \( f(z_0) = z \) for all \( z \in H \), \[ g'(z_0) = \lim_{z \to f(z_0)} \frac{g(z) - g(f(z_0))}{z - f(z_0)} = \lim_{z \to f(z_0)} \frac{1}{f'(s(z_0))} \cdot \frac{g(f(z_0)) - g(z)}{g'(z_0)} \] \[ = \lim_{z \to f(z_0)} \frac{g(f(z_0)) - g(z)}{g'(z_0)(z - f(z_0))} \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 38 Context: # 2 Differentiation Now define \( z_0 = g(x_0) \) and set \[ \phi(w) = \begin{cases} \frac{f(w) - f(w_0)}{w - w_0} & \text{if } w \neq w_0 \\ f'(w_0) & \text{if } w = w_0 \end{cases} \] This is continuous at \( w_0 \) and \(\lim_{w \to w_0} \phi(w) = f'(w_0)\), so we can apply Proposition 2.6: \[ g'(x_0) = \lim_{h \to 0} \frac{1}{\phi(g(x_0 + h))} \cdot \frac{1}{f'(w_0)} \] ## 2.3 The Cauchy–Riemann Equations When considering a real-valued function \( f: \mathbb{R}^2 \to \mathbb{R} \) of two variables, there is no notion of the derivative of a function. For such a function, we instead only have partial derivatives \( \frac{\partial f}{\partial x}(x_0, y_0) \) and \( \frac{\partial f}{\partial y}(x_0, y_0) \) (and also directional derivatives) which depend on the way in which we approach a point \( (x_0, y_0) \in \mathbb{R}^2 \). For a complex-valued function \( f(z) \), we now have a new concept of the derivative \( f'(z_0) \), which definition cannot depend on the way in which we approach a point \( z_0 = (x_0, y_0) \in \mathbb{C} \). It is logical, then, that there should be a relationship between the complex derivative \( f'(z_0) \) and the partial derivatives \[ \frac{\partial f}{\partial x}(z_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0, y_0)}{h} \] and \[ \frac{\partial f}{\partial y}(z_0) = \lim_{k \to 0} \frac{f(x_0, y_0 + k) - f(x_0, y_0)}{k} \] (this definition is exactly as in the real-valued case). This relationship between the complex derivative and partial derivatives is very strong, and it is a powerful computational tool. It is described by the Cauchy–Riemann equations, named after Augustin Louis Cauchy (1789–1857) and Georg Friedrich Bernhard Riemann (1826–1866), even though the equations appeared already in the works of Jean le Rond d'Alembert (1717–1783) and Leonhard Euler (1707–1783). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 39 Context: # THE CAUCHY–RIEMANN EQUATIONS Theorem 2.13. (a) Suppose \( f \) is differentiable at \( z_0 = x_0 + iy_0 \). Then the partial derivatives \( f_x \) and \( f_y \) exist and satisfy \[ \frac{\partial f}{\partial x}(z_0) = \frac{\partial f}{\partial y}(z_0) \cdot i. \tag{2.2} \] (b) Suppose \( f \) is a complex function such that the partial derivatives \( f_x \) and \( f_y \) exist in an open disk centered at \( z_0 \) and are continuous at \( z_0 \). If these partial derivatives satisfy (2.2), then \( f \) is differentiable at \( z_0 \). In both cases (a) and (b), \( f' \) is given by \[ f'(z_0) = \frac{\partial f}{\partial x}(z_0). \] Before proving Theorem 2.13, we note several comments and give two applications. It is traditional, and often convenient, to write the function \( f \) in terms of its real and imaginary parts. That is, we write \( f(z) = f(x, y) = u(x, y) + iv(x, y) \), where \( u \) is the real part of \( f \) and \( v \) is the imaginary part. Then, using the usual shorthand \[ f_x = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \quad \text{and} \quad f_y = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} = v_y - i u_y. \] With this terminology we can rewrite (2.2) as the pair of equations \[ u_{x}(x_0, y_0) = v_{y}(x_0, y_0) \tag{2.3} \] \[ v_{x}(x_0, y_0) = -u_{y}(x_0, y_0). \tag{2.4} \] That is, \[ u_{x}(x_0, y_0) + u_{y}(x_0, y_0) = 0. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 40 Context: # Differentiation (and an analogous identity for \( x \)). Functions with continuous second partials satisfying this partial differential equation on a region \( G \subset \mathbb{C} \) (though not necessarily (2.3)) are called harmonic on \( G \); we will study such functions in Chapter 6. Again, as we will see later, if \( f \) is holomorphic in an open set then the partials of any order of \( u \) and \( v \) exist; hence we will show that the real and imaginary parts of a function that is holomorphic in an open set are harmonic on that set. ## Example 2.14 We revisit Example 2.7 and again consider \( f: \mathbb{C} \to \mathbb{C} \) given by \[ f(z) = z^2 = (x + iy)^2 = (x^2 - 3y^2) + i(3x^2 - y^2). \] Thus, \[ f_x(z) = 3x^2 - 3y^2 + 6ixy \] and \[ f_y(z) = -6xy - 3y^2 - 3ix^2 \] are continuous on \( C \) and satisfy \( f_x = \overline{f_y} \). Thus by Theorem 2.13(b), \( f(z) = z^2 \) is entire. ## Example 2.15 Revisiting Example 2.8 (you saw that coming, didn't you?), we consider \( f: \mathbb{C} \to \mathbb{C} \) given by \[ f(z) = (z^2) = (x - iy)^2 = x^2 - y^2 - 2iy. \] Now, \[ f_x(z) = 2x - 2iy \] and \[ f_y(z) = -2y - 2ix, \] which satisfy \( f_y = \overline{f_x} \) only when \( z = 0 \). (The contrapositive of Theorem 2.13(a)) thus implies that \( f(z) = (z^2) \) is not differentiable on \( C \setminus \{0\} \). ### Proof of Theorem 2.13 (a) If \( f \) is differentiable at \( c = (x_0, y_0) \) then \[ f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}. \] As we know by now, we must get the same result if we restrict \( z \) to be on the real axis and if we restrict it to be on the imaginary axis. In the first case, let \( h = \Delta x \) and \[ f'(x_0) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0)}{\Delta x} \frac{\partial f}{\partial x}(x_0, y_0). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 41 Context: In the second case, \( \Delta z = i \Delta y \) and \[ f'(z_0) = \lim_{y_0 \to y_0} \frac{f(x_0 + i y) - f(x_0)}{i y} = \lim_{y_0 \to 0} \frac{f(x_0 + i y) - f(x_0)}{y} \] Thus, we have shown that \( f'(z_0) = f_x(z_0) - i f_y(z_0) \). (b) Suppose the Cauchy-Riemann equation (2.2) holds and the partial derivatives \( f_x \) and \( f_y \) are continuous in an open disk centered at \( z_0 \). Our goal is to prove that \( f'(z_0) = f_x(z_0) + i f_y(z_0) \). By (2.2), \[ f(z_0) = x_0 + i y_0 \quad f(z_0 + \Delta z) = f(z_0 + \Delta x + i \Delta y) \] On the other hand, we can rewrite the difference quotient for \( f'(z_0) \) as \[ f(z_0 + \Delta z) - f(z_0) = \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} = \frac{f(z_0 + \Delta x + i \Delta y) - f(z_0)}{\Delta z} = \frac{f(z_0 + \Delta x) - f(z_0)}{\Delta z} + \frac{f(z_0 + \Delta y) - f(z_0)}{\Delta z} \] Thus, \[ \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z} = \lim_{\Delta y \to 0} \frac{f(z_0 + i \Delta y) - f(z_0)}{\Delta y} = \lim_{\Delta x \to 0} \left( \frac{f(z_0 + \Delta x + i \Delta y) - f(z_0)}{\Delta z} \right). \] We claim that both limits on the right-hand side are 0, so we have achieved our set goal. The fractions \( \frac{\Delta y}{\Delta z} \) and \( \frac{\Delta x}{\Delta z} \) are bounded in absolute value by 1, so we just need to see that the differences in parentheses are 0. The second term on the right-hand side of (2.5) has a limit of 0 since, by definition, \[ f(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} \] and taking the limit here as \( \Delta z \to 0 \) is the same as taking the limit as \( \Delta x \to 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 42 Context: ```markdown 36 # Differentiation We cannot do something equivalent for the first term in (2.5), since now both \( \Delta x \) and \( \Delta y \) are involved, and both change as \( \Delta x \to 0 \). Instead, we apply the Mean-Value Theorem A.2 for real functions\(^1\) to the real and imaginary parts \( u(x, y) \) and \( v(x, y) \) of \( f(z) \). Theorem A.2 gives real numbers \( \xi \) and \( \eta \) such that \[ \frac{u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0)}{\Delta y} = u_y(x_0, y_0 + \eta) \] and \[ \frac{v(x_0 + \Delta x, y_0 + \Delta y) - v(x_0, y_0)}{\Delta y} = v_y(x_0, y_0 + \xi). \] Thus, \[ \frac{f(x_0 + \Delta x + i \Delta y) - f(x_0 + i y)}{\Delta y} = \frac{u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0)}{\Delta y} - iv_x(x_0, y_0) \] \[ = \left( \frac{u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0)}{\Delta y} \right) - i v_y(x_0, y_0) + \left( \frac{u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0)}{\Delta y} - iv_y(x_0, y_0) \right) = u_y(x_0, y_0) - iv_y(x_0, y_0). \tag{2.6} \] Because \( u \) and \( v \) are continuous at \( (x_0, y_0) \), \[ \lim_{\Delta y \to 0} u(x_0 + \Delta x, y_0 + \Delta y) = u(x_0, y_0) \] and \[ \lim_{\Delta y \to 0} v(x_0 + \Delta x, y_0 + \Delta y) = v(x_0, y_0), \] and so (2.6) goes to \( \Delta z \to 0 \), which we set out to prove. ## 2.4 Constant Functions As a simple application of the definition of the derivative of a complex function, we consider functions that have a derivative of 0. In a typical calculus course, one of the first applications of the Mean-Value Theorem for real-valued functions (Theorem A.1) is to constant functions. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 43 Context: # CONSTANT FUNCTIONS ## Proposition 2.16 If \( I \) is an interval and \( f : I \rightarrow \mathbb{R} \) is a real-valued function with \( f'(x) \) defined and equal to 0 for all \( x \in I \), then there is a constant \( c \in \mathbb{R} \) such that \( f(x) = c \) for all \( x \in I \). **Proof:** The Mean-Value Theorem A.2 says that for any \( x, y \in I \), \[ f(y) - f(x) = f' \left( c \right) (y - x) \] for some \( 0 < a < 1 \). Now \( f' (x + a(y - x)) = 0 \), so the above equation yields \( f(y) = f(x) \). Since this is true for any \( x, y \in I \), the function \( f \) must be constant on \( I \). We do not (yet) have a complex version of the Mean-Value Theorem, and so we will use a different argument to prove that a complex function whose derivative is always 0 must be constant. Our proof of Proposition 2.16 required two key features of the function \( f \), both of which are somewhat obviously necessary. The first is that \( f \) be differentiable everywhere in its domain. In fact, if \( f \) is not differentiable everywhere, we can construct functions that have zero derivative almost everywhere but that have infinitely many values in their image. The second key feature is that the interval \( I \) is connected. It is certainly important for the domain to be connected in both the real and complex cases. For instance, if we define the function \[ f(z) = \begin{cases} 1 & \text{if } \text{Re}(z) \geq 0 \\ 2 & \text{if } \text{Re}(z) < 0 \end{cases} \] then \( f(z) \) is 0 for all \( z \in I \), but \( f \) is not constant. This may seem like a silly example, but it illustrates a pivotal to proving a function is constant that we must be careful of. Recall that a region of \( \mathbb{C} \) is an open connected subset. ## Theorem 2.17 If \( G \subset \mathbb{C} \) is a region and \( f : G \rightarrow \mathbb{C} \) is a complex-valued function with \( f'(z) \) defined and equal to 0 for all \( z \in G \), then \( f \) is constant. **Proof:** We will show that \( f \) is constant along horizontal segments and along vertical segments in \( G \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 44 Context: # Differentiation Suppose that \( H \) is a horizontal line segment in \( G \). Thus there is some number \( y_0 \in \mathbb{R} \) such that the imaginary part of any \( z \in H \) is \( y_0 \). Now consider the real part \( u(x) \) of the function \( f \) for \( x \in H \). Since \( \text{Im}(z) = y_0 \) is constant on \( H \), we can consider \( f(z) = u(x) + iv(y) \) to be just a function of \( x \) for \( z = x + iy \). By assumption, \( f(z) = 0 \) for \( z \in H \) we have \( u(x) = \text{Re}(f(z)) = 0 \). Thus, by Proposition 2.16, \( u(x) \) is constant on \( H \). We can argue the same way to see that the imaginary part \( v(y) \) of \( f(z) \) is constant on \( H \), since \( v(y) = \text{Im}(f(z)) = 0 \) on \( H \). Since both the real and imaginary parts of \( f(z) \) are constant on \( H \), the function \( f(z) \) itself is constant on \( H \). This same argument works for vertical segments, interchanging the roles of the real and imaginary parts. We have thus proved that if \( f \) is constant along horizontal segments and along vertical segments in \( G \), and if \( x \) and \( y \) are two points in \( G \) that can be connected by a path composed of horizontal and vertical segments, we conclude that \( f(x) = f(y) \). But any two points of a region may be connected by finitely many such segments by Theorem 1.12, so \( f \) has the same value at any two points of \( G \), thus proving the theorem. There are a number of surprising applications of Theorem 2.17; see, e.g., Exercises 2.20 and 2.21. ## Exercises 1. Use the definition of limit to show for any \( z \in G \) that \( \lim_{z \to z_0} (az + b) = az_0 + b \). 2. Evaluate the following limits or explain why they don't exist. (a) \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \) (b) \( \lim_{x \to 1} \ln(x + (2x + y)) \) 3. Prove that, if a limit exists, then it is unique. 4. Prove Proposition 2.4. 5. Let \( G \subset \mathbb{C} \) and suppose \( z_0 \) is an accumulation point of \( G \). Show that \( \lim_{z \to z_0} f(z) = 0 \) if and only if \( \lim_{z \to z_0} |f(z)| = 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 45 Context: # CONSTANT FUNCTIONS ## 2.6. Proposition 2.2 is useful for showing that limits do not exist, but it is not at all useful for showing that a limit does exist. For example, define $$ f(x,y) = \frac{xy}{x^2 + y^2} \quad \text{where } (x,y) \neq (0,0). $$ Show that the limits of \( f \) along all straight lines through the origin exist and are equal, but \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist. (Hint: Consider the limit along the parabola \( y = x^2 \).) ## 2.7. Suppose that \( f(z) = u(x,y) + iv(x,y) \) and \( u_0 = u(x_0,y_0) \) and \( v_0 = v(x_0,y_0) \). Prove that $$ \lim_{(x,y) \to (x_0,y_0)} f(z) = u_0 + iv_0 $$ if and only if $$ \lim_{(x,y) \to (x_0,y_0)} u'(x,y) = u_0 \quad \text{and} \quad \lim_{(x,y) \to (x_0,y_0)} v'(x,y) = v_0. $$ ## 2.8. Show that the function \( f : C \to \mathbb{C} \) given by \( f(z) = z^2 \) is continuous on \( C \). ## 2.9. Show that the function \( g : C \to \mathbb{C} \) given by $$ g(z) = \begin{cases} \frac{1}{z} & \text{if } z \neq 0 \\ 1 & \text{if } z = 0 \end{cases} $$ is continuous on \( C \setminus \{0\} \). ## 2.10. Determine where each of the following functions \( f : C \to C \) is continuous: (a) $$ f(z) = \begin{cases} 0 & \text{if } z = 0 \text{ or } |z| \text{ is irrational} \\ 1 & \text{if } |z| \in \mathbb{Q} \setminus \{0\} \end{cases} $$ (b) $$ f(z) = \begin{cases} 0 & \text{if } z = 0 \\ \sin z & \text{if } z \neq 0 \text{ and } z \neq re^{i\theta} \end{cases} $$ #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 46 Context: ``` ## 2.11. Show that the two definitions of continuity in Section 2.1 are equivalent. Consider separately the cases where \(x_0\) is an accumulation point of \(G\) and where \(x_0\) is an isolated point of \(G\). ## 2.12. Consider the function \(f : C \setminus \{0\} \to G\) given by \(f(z) = \frac{1}{z}\). Apply the definition of the derivative to give a direct proof that \(f'(z) = -\frac{1}{z^2}\). ## 2.13. Prove Proposition 2.6. ## 2.14. Prove Proposition 2.10. ## 2.15. Find the derivative of the function \(T(z) = \frac{a z^2 + b z + c}{e^{z}}\), where \(a, b, c \in C\) with \(ad - bc = 0\). When is \(T'(z) = 0\)? ## 2.16. Prove that if \(f(z)\) is given by a polynomial in \(z\) then \(f\) is entire. What can you say if \(f(z)\) is given by a polynomial in \(z = x + iy\) and \(y = \Im z\)? ## 2.17. Prove or find a counterexample. If \(x\) and \(y\) are real valued and continuous, then \(f(z) = u(x,y) + iv(x,y)\) is continuous; if \(u\) and \(v\) are (real) differentiable then \(f\) is (complex) differentiable. ## 2.18. Where are the following functions differentiable? Where are they holomorphic? Determine their derivatives at points where they are differentiable. 1. \(f(z) = e^{z} \cdot e^{c+iy}\) 2. \(f(z) = 2x + iy\) 3. \(f(z) = x^3 + iy\) 4. \(f(z) = e^{c-iy}\) 5. \(f(z) = \cosh x - i \sin y\) 6. \(f(z) = \Im z\) 7. \(f(z) = |z|^2 = x^2 + y^2\) 8. \(f(z) = z^2\) 9. \(f(z) = \Im z\) 10. \(f(z) = \frac{4}{Re z} - i(Re z)^2\) ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 47 Context: ```markdown # Constant Functions ## 2.19 Define \( f(z) = 0 \) if \( \text{Re}(z) - \text{Im}(z) = 0 \), and \( f(z) = 1 \) if \( \text{Re}(z) - \text{Im}(z) \neq 0 \). Show that \( f \) satisfies the Cauchy–Riemann equation (2.2) at \( z = 0 \), yet \( f \) is not differentiable at \( z = 0 \). Why doesn't this contradict Theorem 2.13(b)? ## 2.20 Prove: If \( f \) is holomorphic in the region \( G \subset \mathbb{C} \) and takes real-valued, then \( f' \) is constant in \( G \). (Hint: Use the Cauchy–Riemann equations (2.3) to show that \( f' = 0 \).) ## 2.21 Prove: If \( f(z) \) and \( \overline{f}(z) \) are both holomorphic in the region \( G \subset \mathbb{C} \), then \( f(z) \) is constant in \( G \). ## 2.22 Suppose \( f \) is entire and can be written as \( f(z) = u(x,y) + iv(x,y) \), that is, the real part of \( f \) depends only on \( x \) and the imaginary part of \( f \) depends only on \( y \). Assume that \( f(z) = az + b \) for some \( a \in \mathbb{R} \) and \( b \in \mathbb{C} \). ## 2.23 Suppose \( f \) is entire, with real and imaginary parts \( u \) and \( v \) satisfying \[ u(x,y) v(x,y) = 3 \] for all \( z = x + iy \). Show that \( f \) is constant. ## 2.24 Prove that the Cauchy–Riemann equations take on the following form in polar coordinates: \[ \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text{and} \quad \frac{1}{r} \frac{\partial v}{\partial r} = -\frac{\partial u}{\partial \theta}. \] ## 2.25 For each of the following functions, find a function \( f \) such that \( u + iv \) is holomorphic in some region. Maximize that region. (a) \( u(x,y) = x^2 - y^2 \) (b) \( u(x,y) = \cosh(y) \) (c) \( u(x,y) = 2x^2 + x + 1 - 2y^2 \) (d) \( u(x,y) = \frac{xy}{x^2 + y^2} \) ## 2.26 Is \( u(x,y) = \frac{x}{x^2+y^2} \) harmonic on \( \mathbb{C} \)? What about \( v(x,y) = \frac{y}{x^2+y^2} \)? ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 48 Context: ``` 42 # Differentiation ## 2.27 Consider the general real homogeneous quadratic function \( u(x,y) = ax^2 + bxy + cy^2 \), where \( a, b, \) and \( c \) are real constants. (a) Show that \( u \) is harmonic if and only if \( a + c = 0 \). (b) If \( u \) is harmonic then show that \( u \) is the real part of a function of the form \( f(z) = A z^2 \) for some \( A \in \mathbb{C} \). Give a formula for \( A \) in terms of \( a \) and \( c \). ## 2.28 Re-prove Proposition 2.10 by using the formula for \( f' \) given in Theorem 2.13. ## 2.29 Prove that, if \( G \subset \mathbb{C} \) is a region and \( f : G \to \mathbb{C} \) is a complex-valued function with \( f''(z) \) defined and equal to 0 for all \( z \in G \), then \( f(z) = az + b \) for some \( a, b \in \mathbb{C} \). (Hint: Use Theorem 2.17 to show that \( f'(z) = a \), and then use Theorem 2.17 again for the function \( f(z) - az \).) ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 49 Context: # Chapter 3 ## Examples of Functions _To many, mathematics is a collection of theorems. For us, mathematics is a collection of examples; a theorem is a statement about a collection of examples and the purpose of proving theorems is to classify and explain the examples..._ — John B. Conway In this chapter we develop a toolkit of complex functions. Our ingredients are familiar from calculus: linear functions, exponentials and logarithms, and trigonometric functions. Yet, when we move these functions into the complex world, they take on—at times drastically different—new features. ### 3.1 Möbius Transformations The first class of functions that we will discuss in some detail are built from linear polynomials. **Definition.** A linear fractional transformation is a function of the form \[ f(z) = \frac{az + b}{cz + d} \] where \(a, b, c, d \in \mathbb{C}\) and \(ad - bc \neq 0\). Then \(f\) is called a Möbius transformation. Exercise 2.16 said that any polynomial \(p\) is an entire function, and so the linear fractional transformation \(f(z) = \frac{az + b}{cz + d}\) is holomorphic on \(\mathbb{C} \setminus \{-\frac{d}{c}\}\), unless \(c = 0\) (in which case \(c\) is entire). If \(c \neq 0\) then \(f(z) = \frac{az + b}{cz + d}\) implies \(ad - bc \neq 0\), which means that a Möbius transformation \(f(z) = \frac{az + b}{cz + d}\) will never take on the value \(-\frac{b}{d}\). Our first proposition in this chapter says that with these small observations about the domain and image of a Möbius transformation, we obtain a class of bijections, which are quite special among complex functions. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 50 Context: # EXAMPLES OF FUNCTIONS ## Proposition 3.1 Let \( a, b, c, d \in \mathbb{C} \) with \( c \neq 0 \). Then \( f : C \setminus \{ -\frac{d}{c} \} \to C \setminus \{ \infty \} \) given by \[ f(z) = \frac{az + b}{cz + d} \] has the inverse function \( f^{-1} : C \setminus \{ \infty \} \to C \setminus \{ -\frac{d}{c} \} \) given by \[ f^{-1}(w) = \frac{dz - b}{-cz + a}. \] We remark that the same formula for \( f^{-1} \) works when \( c = 0 \), except that in this case both domain and image of \( f \) are \( C \): see Exercise 3.2. In either case, we note that the inverse of a Möbius transformation is another Möbius transformation. ## Example 3.2 Consider the linear fractional transformation \( f(z) = \frac{1}{z} \). This is a Möbius transformation (check the condition!) with domain \( C \setminus \{ 0 \} \) whose inverse can be computed via \[ \frac{z - 1}{i + z} = w \quad \Longleftrightarrow \quad z = \frac{iw + 1}{-iw + 1} \] so that \( f^{-1}(w) = \frac{-1}{w} \) with domain \( C \setminus \{ -i \} \). ## Proof of Proposition 3.1 We first prove that \( f \) is one-to-one. If \( f(z_1) = f(z_2) \), that is, \[ \frac{az_1 + b}{cz_1 + d} = \frac{az_2 + b}{cz_2 + d}, \] then \[ (ad - bc)(z_2 - z_1) = 0. \] Since \( ad - bc \) implies that \( z_1 = z_2 \). This shows that \( f \) is one-to-one. ## Exercise 3.1 Verifies that the Möbius transformation \( g(z) = \frac{az + b}{cz + d} \) is the inverse of \( f \), and by what we have just proved, \( g \) is also one-to-one. But this implies that \( f \circ f^{-1} : C \setminus \{ \infty \} \to C \setminus \{ -\frac{d}{c} \} \) is onto. We remark that Möbius transformations provide an immediate application of Proposition 2.11, as the derivative of \( f(z) = \frac{az + b}{cz + d} \) is \[ f'(z) = \frac{d(cz + d) - c(az + b)}{(cz + d)^2} = \frac{ad - bc}{(cz + d)^2} \] and thus never zero. Proposition 2.11 implies that Möbius transformations are conformal, that is, they preserve angles. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 51 Context: # Mӧbius Transformations Mӧbius transformations have even more fascinating geometric properties. En route to an example of such, we introduce some terminology. Special cases of Mӧbius transformations are translations \( f(z) = z + b \), dilations \( f(z) = az \), and inversions \( f(z) = \frac{1}{z} \). The next result says that if we understand those three special Mӧbius transformations, we understand them all. ## Proposition 3.3 Suppose \( f(z) = \frac{az+b}{cz+d} \) is a linear fractional transformation. If \( c = 0 \), then \[ f(z) = \frac{a}{d} z + \frac{b}{d} \] and if \( c \neq 0 \), then \[ f(z) = \frac{bz - ad}{cz} + \frac{1}{c} \]. In particular, every linear fractional transformation is a composition of translations, dilations, and inversions. **Proof.** Simplify. ## Theorem 3.4 Mӧbius transformations map circles and lines into circles and lines. ## Example 3.5 Continuing Example 3.2, consider \( f(z) = \frac{1}{z} \). For \( z \in \mathbb{R} \), \[ f(\overline{z}) = \frac{e^{-1}}{i \overline{z} + 1} = \frac{(e^{-1}-1)(\overline{z} + 1)}{|e| + 1|^2} = \frac{e^{-c}}{|e| + 1|^2} \cdot \frac{2lm(e)}{|e| + 1|^2} = \frac{2sinp}{|e| + 1|^2}, \] which is a real number. Thus, Theorem 3.4 implies that \( f \) maps the unit circle to the real line. **Proof of Theorem 3.4:** Transformations and dilations certainly map circles and lines into circles and lines, so by Proposition 3.3, we only have to prove the statement for the transformation for the inversion \( f(z) = \frac{1}{z} \). The equation for a circle centered at \( x_0 + iy_0 \) with radius \( r \) can be transformed to \[ a(x^2 + y^2) + \beta x + \gamma y + 8 = 0 \quad (3.1). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 52 Context: # Examples of Functions For real numbers \( a, b, \gamma, y, r \) and \( s \) that satisfy \( \beta^2 + \gamma^2 > 4\alpha \) (Exercise 3.3). The form (3.1) is more convenient for us, because it includes the possibility that the equation describes a line (precisely when \( a = 0 \)). Suppose \( x + iy \) satisfies (3.1); we need to prove that \( x + iy = \frac{x - iy}{x^2 + y^2} \) satisfies a similar equation. \[ u + iv = \frac{x - iy}{x^2 + y^2}. \] We can rewrite (3.1) as: \[ 0 = \alpha + \frac{y}{x^2 + y^2} + \frac{8}{x^2 + y^2}. \] This leads to: \[ 0 = \alpha + \beta u + \gamma (a + b + r^2). \tag{3.2} \] But this equation, in conjunction with Exercise 3.3, says that \( x + iy \) lies on a circle or line. ## 3.2 Infinity and the Cross Ratio In the context of Möbius transformations, it is useful to introduce a formal way of saying that a function \( f \) "blows up" in absolute value, and this gives rise to a notion of infinity. **Definition.** Suppose \( f : G \to \mathbb{C} \). 1. \( \lim_{x \to a} f(x) = \infty \) means that for every \( M > 0 \) we can find \( \delta > 0 \) so that, for all \( x \in G \) satisfying \( 0 < |x - a| < \delta \), we have \( |f(x)| > M \). 2. \( \lim_{x \to b} f(x) = L \) means that for every \( \epsilon > 0 \) we can find \( N > 0 \) so that, for all \( x \in G \) satisfying \( |x| > N \), we have \( |f(x) - L| < \epsilon \). 3. \( \lim_{x \to \infty} f(x) = \infty \) means that for every \( M > 0 \) we can find \( N > 0 \) so that, for all \( x \in G \) satisfying \( |x| > N \), we have \( |f(x)| > M \). In the first definition we require that \( a \) be an accumulation point of \( G \), while in the second and third we require that \( b \) be an "extended accumulation point" of \( G \), in the sense that for every \( B > 0 \) there is some \( z \in G \) with \( |z| > B \). As in Section 2.1, the limit, in any of these senses, is unique if it exists. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 53 Context: # INFINITY AND THE CROSS RATIO ### Example 3.6 We claim that \(\lim_{z \to \infty} f(z) = \infty\): Given \(M > 0\), let \(\delta = \frac{1}{M}\). Then \(0 < |z| < \delta\) implies \[ |f(z)| = \left| \frac{1}{z} \right| = \frac{1}{|z|} > M. \] ### Example 3.7 Let \(f(z) = \frac{az + b}{cz + d}\) be a Möbius transformation with \(c \neq 0\). Then \(\lim_{z \to \infty} f(z) = \frac{a}{c}\). To simplify the notation, introduce the constant \(L = |az - b| \geq |c| |z| - |d|.\) Given \(\epsilon > 0\), let \(N = \frac{|d|}{|a| - |c|}\). Then \(|z| > N\) implies, with the reverse triangle inequality (Corollary 1.7(b)), \[ |z + d| \geq ||c| z - |d| \geq |c| |z| - |d| \] and so \[ f(z) - \frac{a}{c} = \frac{(az + b) - \frac{a}{c}(cz + d)}{cz + d} = \frac{(ad - bc)}{c(cz + d)} \cdot L < \epsilon. \] We stress that \(\infty\) is not a number in \(\mathbb{C}\), just as \(0\) are not numbers in \(\mathbb{R}\). However, we can extend \(\mathbb{C}\) by adding on \(\infty\) if we are careful. We do so by realizing that we are always talking about a limit when handling \(\infty\). It turns out (Exercise 3.11) that the usual limit rules behave well when we mix complex numbers and zero. For example, if \(\lim_{z \to c} f(z) = L\) and \(\lim_{z \to c} g(z) = \infty\), then the usual limit of \(f(z) + g(z)\) still holds and gives \[ \lim_{z \to c} (f(z) + g(z)) = \infty. \] The following definition captures the philosophy of this paragraph. ### Definition The extended complex plane is the set \(\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}\), together with the following algebraic properties: - (a) \(0 + \infty = \infty\) - (b) \(x + \infty = \infty\) for all \(x \in \mathbb{C}\) - (c) \(0 \cdot \infty = 0\) - (d) \(\frac{a}{0} = \infty\) - (e) if \(a \neq 0\) then \(\frac{0}{a} = 0\). The extended complex plane is also called the Riemann sphere or the complex projective line, denoted \(\text{CP}^1\). If a calculation involving \(\infty\) is not covered by the rules above then we must investigate the limit more carefully. For example, it may seem strange that \(0 + \infty\) is defined as \(\infty\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 54 Context: # EXAMPLES OF FUNCTIONS is not defined, but if we take the limit of \( z + (-z) = 0 \) as \( z \to \infty \) we will get \( 0 \), even though the individual limits of \( z \) and \( -z \) are both \( \infty \). Now we reconsider Möbius transformations with \( c \) in mind. For example, \( f(z) = \frac{az + b}{cz + d} \) is now defined for \( z = 0 \) and \( z = \infty \), with \( f(0) = \frac{b}{d} \) and \( f(\infty) = \frac{a}{c} \), so we might argue the proper domain for \( f(z) \) is actually \( \mathbb{C} \). Let’s consider the other basic types of Möbius transformations. A translation \( f(z) = z + b \) is now defined for \( z = \infty \), with \( f(\infty) = \infty \) and a dilation \( f(z) = az \) (with \( a \neq 0 \)) is also defined for \( z = \infty \), with \( f(\infty) = \infty \). Since every Möbius transformation can be expressed as a composition of translations, dilations, and inversions (Proposition 3.3), we see that every Möbius transformation may be interpreted as a transformation of \( \mathbb{C} \) onto \( \mathbb{C} \). This general case is summarized in the following extension of Proposition 3.1. ## Corollary 3.8 Suppose \( a, b, c, d \) are \( \neq 0 \) and consider \( f: \mathbb{C} \to \mathbb{C} \) defined through \[ f(z) = \begin{cases} \frac{az + b}{cz + d} & \text{if } z \in \mathbb{C} \setminus \{-\frac{d}{c}\} \\ \infty & \text{if } z = -\frac{d}{c} \\ \end{cases} \] Then \( f \) is a bijection. This corollary also holds for \( c = 0 \) if we define \( f(\infty) = 0 \). ## Example 3.9 Continuing Examples 3.2 and 3.5, consider once more the Möbius transformation \( f(z) = \frac{1}{z} \). With the definitions \( f(1) = 1 \) and \( f(\infty) = 0 \), we can extend \( f \) to a function \( \mathbb{C} \). With \( 0 \) on our mind, we can also add some insight to Theorem 3.4. We recall that in Example 3.5, we proved that \( f(z) = \frac{1}{z} \) maps the unit circle to the real line. Essentially, the same proof shows that, more generally, any circle passing through \( \infty \) gets mapped to a line (see Exercise 3.4). The original domain of \( f \) was \( \mathbb{C} \setminus \{0\} \), so the point \( z = 1 \) must be excluded from these circles. However, by thinking of \( f \) as a function from \( \mathbb{C} \) to \( \mathbb{C} \), we can put \( z = 1 \) back into the picture, and \( f \) transforms the circle defined by \( |z| = 1 \) into a line plus a point. If we make this a definition, then Theorem 3.4 can be expressed as: any Möbius transformation \( f(z) \) transforms circles to lines. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 55 Context: We can take this remark a step further, based on the idea that three distinct points in \( \mathbb{C} \) determine a unique circle passing through them. If the three points are in \( \mathbb{C} \) and non-collinear, this fact comes straight from Euclidean geometry: if the three points are on a straight line or if one of the points is at infinity, then the circle is a straight line plus one. ### Example 3.10 The Möbius transformation \( f : \mathbb{C} \to \mathbb{C} \) given by \( f(z) = \frac{az + b}{cz + d} \) maps: - \( 1 \to 0 \) - \( i \to 1 \) - \( -1 \to \infty \) The points \( 1, i, -1 \) uniquely determine the unit circle and the points \( 0, 1, \) and \( \infty \) uniquely determine the real line, viewed as a circle in \( \mathbb{C} \). Thus Corollary 3.8 implies that if \( f \) maps the unit circle to \( \mathbb{R} \), which we already concluded in Example 3.5. Conversely, if we know where three distinct points in \( \mathbb{C} \) are transformed by a Möbius transformation, then we should be able to figure out everything about the transformation. There is a computational device that makes this easier to see. ### Definition If \( z_1, z_2, z_3 \) are any four points in \( \mathbb{C} \) with \( z_1, z_2, \) and \( z_3 \) distinct, then their cross ratio is defined as: \[ [z_1, z_2, z_3, z_4] = \frac{(z_1 - z_2)(z_3 - z_4)}{(z_1 - z_4)(z_3 - z_2)} \] This includes the implicit definitions \([z_1, z_2, z_3, \infty] = 0\) and, if one of \( z_1, z_2, \) or \( z_3 \) is \( 0 \), then the two terms containing \( 0 \) are canceled; e.g., \([0, z_2, z_3, \infty] = \frac{z_2 - z_3}{0 - z_2} = \frac{1}{-1}\). ### Example 3.11 Our running example \( f(z) = \frac{1}{z} \) can be written as \( f(z) = [1, i, -1] \). ### Proposition 3.12 The function \( f : \mathbb{C} \to \mathbb{C} \) defined by \( f(z) = [z_1, z_2, z_3, z] \) is a Möbius transformation that satisfies: \[ f(z_1) = 0, \quad f(z_2) = 1, \quad f(z_3) = \infty. \] Moreover, if any Möbius transformation with \( g(z_1) = 0, \, g(z_2) = 1, \) and \( g(z_3) = \infty \), then \( f \) and \( g \) are identical. ### Proof Most of this follows from our definition of \( f \), but we need to prove the uniqueness statement. By Proposition 3.11, the inverse \( f^{-1} \) is a Möbius transformation and, by Exercise 3.10, the composition \( g = f \circ f^{-1} \) is again a Möbius transformation. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 56 Context: # Examples of Functions Transformation. Note that \( h(0) = g(f^{-1}(0)) = g(0) = 0 \) and, similarly, \( h(1) = 1 \) and \( h(\infty) = \infty \). If we write \( h(z) = \frac{az + b}{cz + d} \) then - \( 0 = h(0) = \frac{b}{d} \Rightarrow b = 0 \) - \( \infty = h(\infty) = \frac{a}{c} \Rightarrow c = 0 \) - \( 1 = h(1) = \frac{a + b}{c + d} = \frac{0 + d}{0 + d} \Rightarrow d = a \) and so \[ h(z) = \frac{az + b}{c z + d} \Rightarrow \frac{b}{d} = z, \] the identity function. But since \( b = g \circ f^{-1} \), this means that \( f \) and \( g \) are identical. So if we want to map three given points of \( \mathbb{C} \) to \( 0, 1, \) and \( \infty \) by a Möbius transformation, then the cross ratio gives us the only way to do it. What if we have three points \( z_1, z_2, z_3 \) and we want to map them to three other points \( w_1, w_2, w_3 \)? ## Corollary 3.13 Suppose \( z_1, z_2, z_3 \) are distinct points in \( \mathbb{C} \) and \( w_1, w_2, w_3 \) are distinct points in \( \mathbb{C} \). Then there is a unique Möbius transformation \( h \) satisfying \( h(z_1) = w_1, h(z_2) = w_2, h(z_3) = w_3 \). **Proof:** Let \( z = g^{-1}(f(z)) = [z_1, z_2, z_3] \) and \( g(w) = [w, w_1, w_2, w_3] \). Uniqueness follows as in the proof of Proposition 3.12. ## 3.3 Stereographic Projection The addition of the point at infinity \( \mathbb{C} \) gives the plane a useful structure. This structure is revealed by a famous function called stereographic projection, which gives us a way of visualizing the extended complex plane—that is, with the point at infinity—in \( \mathbb{R}^3 \), as the unit sphere. It also provides a way of seeing that a line in the extended complex plane really is a circle, and of visualizing Möbius functions. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 57 Context: # Stereographic Projection To begin, we think of \(C\) as the \((x,y)\)-plane in \(\mathbb{R}^3\), that is, \(C = \{(x,y,0) \in \mathbb{R}^3\}\). To describe stereographic projection, we will be less concerned with actual complex numbers \(x + iy\) and more concerned with their coordinates. Consider the unit sphere \[ S^2 = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}. \] The sphere and the complex plane intersect in the set \(\{(x,y,0) : x^2 + y^2 = 1\}\), which corresponds to the equator on the sphere and the unit circle on the complex plane, as depicted in Figure 3.1. Let \(N = (0,0,1)\), the north pole of \(S^2\), and let \(S = (0,0,-1)\), the south pole. ![Figure 3.1: Setting up stereographic projection.](#) ## Definition The stereographic projection of \(S^2\) to \(\hat{C}\) from \(N\) is the map \(\varphi: S^2 \setminus \{N\} \to \hat{C}\) defined as follows. For any point \(P \in S^2 \setminus \{N\}\), let \(x\) be the \(x\)-coordinate of \(P\) and let \(l\) be the line through \(N\) and \(P\). The \(z\)-coordinate of \(P\) is strictly less than \(1\), the line through \(N\) and \(P\) intersects \(C\) in exactly one point \(Q\). Define \(\varphi(P) = Q\). We also declare that \(\varphi(N) = \infty\). ## Proposition 3.14 The map \(\varphi\) is given by \[ \varphi(x,y,z) = \left( \frac{x}{1-z}, \frac{y}{1-z}, 0 \right) \quad \text{if } z \neq 1, \] \[ \varphi(x,y,z) = \infty \quad \text{if } z = 1. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 58 Context: # Examples of Functions It is bijective, with inverse map \[ \varphi^{-1}(p,q,0) = \left( \frac{2p}{p^2 + q^2 + 1}, \frac{2q}{p^2 + q^2 + 1}, \frac{p^2 + q^2 - 1}{p^2 + q^2 + 1} \right) \] and \(\varphi^{-1}(\infty) = (0,0,1)\). **Proof:** Given \( P = (x,y,z) \in S^3 \setminus \{N\} \), the straight line through \( N \) and \( P \) is parametrized by \[ r(t) = N + t(P - N) = (0,0,1) + t((x,y,z) - (0,0,1)) = (tx, ty, t(z + 1)). \] where \( t \in \mathbb{R} \). When \( r(t) \) hits \( C \), the third coordinate is \( 0 \), so it must be that \( t = \frac{1}{1-z} \). Plugging this value of \( t \) into the formula for \( r \) yields \( x \) as stated. To see the formula for the inverse map, we begin with a point \( Q = (p,q,0) \in C \) and solve for a point \( P = (x,y,z) \in S^3 \) such that \( \varphi(Q) = P \). The point \( P \) satisfies the equation \( x^2 + y^2 + z^2 = 1 \). The equation \( \varphi(Q) \) tells us that \( z = \frac{p^2 + q^2 - 1}{p^2 + q^2 + 1} \). Thus, we solve three equations for three unknowns. The latter two equations yield: \[ p^2 + q^2 = x^2 + y^2 \cdot \frac{1-x^2}{(1-x^2)} + \frac{1 - z}{1 - z} \] Solving \( p^2 + q^2 = \frac{1}{z^2} \) and then plugging this into the identities \( x = \varphi(1) \) and \( y = \varphi(1) \) proves the desired formula. It is easy to check that \( \varphi \circ \varphi^{-1} \) and \( \varphi^{-1} \circ \varphi \) are both the identity map; see Exercise 3.25. ## Theorem 3.15 The stereographic projection takes the set of circles in \( S^2 \) bijectively to the set of circles in \( C \), where for a circle \( C \subset S^2 \) we have \( \varphi(C) \) that is a line in \( C \) if and only if \( N \in C \). **Proof:** A circle in \( S^2 \) is the intersection of \( S^2 \) with some plane \( H \). If \( (x_0,y_0,z_0) \) is a normal vector for \( H \), then there is a unique point \( k \) so that \( H \) is given by \[ \{(x,y,z) \in \mathbb{R}^3 : (x,y,z) \cdot (x_0,y_0,z_0) = k\}. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 59 Context: # Stereographic Projection By possibly changing \(k\), we may assume that \((x_0, y_0) \in S^2\). We may also assume that \(0 \leq k \leq 1\), since if \(0 < k < 1\) we can replace \((x_0, y_0)\) with \((x_0 - y_0, y_0)\) and if \(k > 1\) then \(H \cap S^2 = \emptyset\). Consider the circle of intersection \(H \cap S^2\). A point \((p_x, p_y)\) in the complex plane lies on the image of this circle under \(\varphi^{-1}(p)\) if and only if \(\varphi^{-1}(p)\) satisfies the defining equation for \(H\). Using the equations from Proposition 3.14 for \(\varphi^{-1}(p)\), we see that \[ \left( \rho - k \right)^2 + \left( (x_0 - p_x) \rho + (y_0 - p_y) \right)^2 = \rho^2 + k. \] If \(k = 0\), this is a straight line in the \((\rho, \phi)\)-plane. Moreover, every line in the \((\rho, \phi)\)-plane can be obtained in this way. Note that \(k = 0\) if and only if \(N \in H\), which is if and only if the image under \(g\) is a straight line. If \(k \neq 0\), then completing the square yields \[ \left( \frac{\rho - x_0}{k} \right)^2 + \left( \frac{\phi - y_0}{k} \right)^2 = 1 - \frac{k^2}{(x_0 - y_0)^2}. \] Depending on whether the right-hand side of this equation is positive, or negative, this is the equation of a circle, point, or the empty set in the \((\rho, \phi)\)-plane, respectively. These three cases happen when \(0 < k < 1\) and \(k = 1\), respectively. Only the first case corresponds to a circle in \(S^2\). Exercise 3.28 verifies that every circle in the \((\rho, \phi)\)-plane arises in this manner. We can now think of the extended complex plane as a sphere in \(\mathbb{R}^2\), the aforementioned Riemann sphere. It is particularly nice to think about the basic Mobius transformations that their effect on the Riemann sphere. We will describe inversions. It is worth thinking about, though beyond the scope of this book, how other Mobius functions behave. For instance, a rotation \(f(z) = e^{i\theta}z\), composed with \(g^{-1}\), can be seen to be a rotation of \(S^2\). We encourage you to verify this and consider the broader implications of visualizing a real dilation, \(f(z) = rf(z_0)\), a rotation, \(f(z) = e^{i\theta}z\), and a dilation in some sense can be understood as a rotation, in that each moves points along sets of circles. We now use stereographic projection to take another look at \(f(z) = z\). We want to know what this function does to the sphere \(S^2\). We will take a point \((x, y, z) \in S^2\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 60 Context: # Examples of Functions Project it to the plane by the stereographic projection \( \pi \), apply \( f \) to the point that results, and then pull this point back to \( S^2 \) by \( \pi^{-1} \). We know \( \varphi(x,y,z) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right) \) which we now regard as the complex number \[ \rho + i \gamma = \frac{-x}{1-z} + i \frac{y}{1-z}. \] We know from a previous calculation that \( p^2 + q^2 = \frac{1}{z^2} \), which gives \( x^2 + y^2 = (1 + z)(1 - z) \). Thus, \[ f\left( \frac{x}{1-z} + i \frac{y}{1-z} \right) = \frac{1 - z}{x + iy} = \frac{1 - z}{(1 - z) x + i (1 - z) y} = \frac{1 - 2X - iY}{x^2 + y^2}. \] Rather than plug this result into the formulas for \( \pi^* \), we can just ask what triple of numbers will be mapped to this particular pair using the formulas \( \varphi(x,y,z) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right) \). Thus, we have shown that the effect of \( f \) on \( S^2 \) is to take \( (x,y,z) \) to \( (x, -y, -z) \). This is a rotation around the \( x \)-axis by 180 degrees. We now have a second argument that \( f(z) \) takes circles and lines to circles and lines. A circle in \( C \) is taken to a circle on \( S^2 \) by \( \pi \). Then \( f(z) = \frac{1}{z} \) rotates the sphere which certainly takes circles to circles. Now \( g \) takes circles back to circles and lines. We can also say that the circles that go to lines under \( f(z) = \frac{1}{z} \) are the circles through 0 because 0 is mapped to \( (0, 0) \) under \( \pi \), and so a circle through 0 in \( C \) goes to a circle through the south pole on \( S^2 \). Now 180-degree rotation about the \( x \)-axis takes the south pole to the north pole, and our circle is now passing through \( N \). But we know that \( g \) will take this circle to a line in \( C \). We end by mentioning that there is, in fact, a way of putting the complex metric on \( S^2 \). It is certainly not the (finite) distance function induced by \( \pi \). Indeed, the origin in the complex plane corresponds to the south pole of \( S^2 \). We have to be able to arbitrarily far away from the origin in \( C \), so the complex distance function has to increasingly grow with the \( z \)-coordinate. The closest points are to the north pole \( N \) corresponding to 0 in \( C \), the larger their distance to the origin, and to each other! In this light, a 'line' in the Riemann sphere \( S^2 \) corresponds to a circle in \( S^2 \) through \( N \). In the regular sphere, the circle has finite lengths, but as a line on the Riemann sphere with the complex metric, it has infinite length. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 61 Context: # Exponential and Trigonometric Functions ## 3.4 Exponential and Trigonometric Functions To define the complex exponential function, we once more borrow concepts from calculus, namely the real exponential \( e^x \) and the real sine and cosine, and we finally make sense of the notation \( e^{x + iy} = e^x (\cos y + i \sin y) \). ### Definition The **complex exponential function** is defined for \( z = x + iy \) as: \[ \exp(z) = e^x (\cos y + i \sin y) = e^z. \] This definition seems a bit arbitrary. Our first justification is that all exponential rules that are used to from real numbers carry over to the complex case. They mainly follow from Proposition 1.3 and are collected in the following. ### Proposition 3.16 For all \( z_1, z_2 \in \mathbb{C} \): (a) \(\exp(z_1) \exp(z_2) = \exp(z_1 + z_2)\) (b) \(\exp(-z) = \frac{1}{\exp(z)}\) (c) \(\exp(iz) \neq 0\) (d) \(\exp(z) = \exp(\text{Re}(z))\) (e) \(\exp(z + 2\pi i) = \exp(z)\) (f) \(\exp(0) = \exp(z)\). Identity (f) is very special and has no counterpart for the real exponential function. It says that the complex exponential function is periodic with period \( 2\pi i \). This has many interesting consequences; one that may not seem too pleasant at first sight is the fact that the complex exponential function is not one-to-one. Identity (f) is not only remarkable, but we invite you to meditate on its proof below; it gives a strong indication that our definition of exp is reasonable. We also note that (f) implies that \( \exp \) is infinite. We leave the proof of Proposition 3.16 as Exercise 3.34 but give one example. ### Proof of (f) The partial derivatives of \( f(z) = \exp(z) \) are: \[ \frac{\partial f}{\partial x} = e^{x + iy} \cos y \quad \text{and} \quad \frac{\partial f}{\partial y} = e^{x + iy} (-\sin y + i \cos y). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 62 Context: # EXAMPLES OF FUNCTIONS They are continuous in \( \mathbb{C} \) and satisfy the Cauchy–Riemann equation (2.2): \[ \frac{\partial f}{\partial x} (z) = \frac{\partial f}{\partial y} (z) \] for all \( z \in \mathbb{C} \). Thus Theorem 2.13 says that \( f(z) = \exp(z) \) is entire with derivative \[ f'(z) = \frac{\partial f}{\partial x} (z) = \exp(z). \] We should make sure that the complex exponential function specializes to the real exponential function for real arguments: indeed, if \( z = x \in \mathbb{R} \), then \[ \exp(x) = e^x = \left( \cos(0) + i\sin(0) \right) = e^x. \] The trigonometric functions—sine, cosine, tangent, cotangent, etc.—also have complex analogues; however, they do not play the same prominent role as in the real case. In fact, we can define them as merely being special combinations of the exponential function. ## Definition The (complex) sine and cosine are defined as: \[ \sin z = \frac{1}{2i} \left( \exp(iz) - \exp(-iz) \right) \] and \[ \cos z = \frac{1}{2} \left( \exp(iz) + \exp(-iz) \right), \] respectively. The tangent and cotangent are defined as: \[ \tan z = \frac{\sin z}{\cos z} = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz}) - 1} \] and \[ \cot z = \frac{\cos z}{\sin z} = \frac{e^{iz} + 1}{e^{iz} - e^{-iz}}, \] respectively. Note that to write tangent and cotangent in terms of the exponential function, we used the fact that \( \exp(iy) - \exp(-iy) = 0 \). Because \( \exp \) is entire, so are sine and cosine. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 63 Context: # Exponential and Trigonometric Functions ![Image properties of the exponential function](path/to/image.png) Figure 3.2: Image properties of the exponential function. As with the exponential function, we should make sure that we're not redefining the real sine and cosine: if \( z \in \mathbb{R} \) then \[ \sin x = \frac{1}{2i} \left( \exp(ix) - \exp(-ix) \right) \] \[ = \frac{1}{2} \left( \cos x + i \sin x - \cos(-x) - i \sin(-x) \right) = \sin x. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 64 Context: # Examples of Functions A similar calculation holds for the cosine. Not too surprisingly, the following properties follow mostly from Proposition 3.16. ## Proposition 3.17. For all \( z_1, z_2 \in \mathbb{C} \): \[ \begin{align*} \sin(-z) &= -\sin z & \cos(-z) &= \cos z \\ \sin(z + 2\pi) &= \sin z & \cos(z + 2\pi) &= \cos z \\ \tan(z + \pi) &= \tan z & \cot(z + \pi) &= \cot z \\ \sin(z + \frac{\pi}{2}) &= \cos z & \cos(z + \frac{\pi}{2}) &= -\sin z \\ \cos(z_1 + z_2) &= \cos z_1 \cos z_2 - \sin z_1 \sin z_2 \\ \sin(z_1 + z_2) &= \sin z_1 \cos z_2 + \cos z_1 \sin z_2 \\ \cos^2 z + \sin^2 z &= 1 & \frac{d}{dz} \sin z &= \cos z \\ \frac{d}{dz} \cos z &= -\sin z \end{align*} \] Finally, we end with caution: unlike in the real case, the complex sine and cosine are not bounded—consider, for example, \(\sin(y)\) as \(y \to \infty\). We end this section with a remark on hyperbolic trigonometric functions. The hyperbolic sine, cosine, tangent, and cotangent are defined as in the real case: ## Definition: \[ \begin{align*} \sinh z &= \frac{1}{2} \left( e^{z} - e^{-z} \right) & \cosh z &= \frac{1}{2} \left( e^{z} + e^{-z} \right) \\ \tanh z &= \frac{\sinh z}{\cosh z} = \frac{e^{z} - e^{-z}}{e^{z} + e^{-z}} & \coth z &= \frac{1}{\tanh z} = \frac{\cosh z}{\sinh z} - 1 \end{align*} \] As such, they are yet more special combinations of the exponential function. They still satisfy the identities you already know, e.g., \[ \frac{d}{dz} \sinh z = \cosh z \quad \text{and} \quad \frac{d}{dz} \cosh z = \sinh z. \] Moreover, they are related to the trigonometric functions via: \[ \sinh(iz) = i \sin z \quad \text{and} \quad \cosh(iz) = \cos z. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 65 Context: # Logarithms and Complex Exponentials The complex logarithm is the first function we'll encounter that is of somewhat tricky nature. It is motivated as an inverse to the exponential function, that is, we're looking for a function \( \Log \) such that \[ \exp(\Log(z)) = z = \Log(\exp z). \] But because \( \exp \) is not one-to-one, this is too much to hope for. In fact, given a function \( \Log \) that satisfies the first equation, the function \( f(z) = \Log(z) + 2\pi i k \) does as well, and so there cannot be an inverse of \( \exp \) (which would have to be unique). On the other hand, \( \exp \) becomes one-to-one if we restrict its domain, so there is hope for a logarithm if we're careful about its construction and about its domain. ## Definition Given a region \( G \), any continuous function \( \Log: G \rightarrow \mathbb{C} \) that satisfies \( \exp(\Log(z)) = z \) is a branch of the logarithm (on \( G \)). To make sure this definition is not vacuous, let's write, as usual, \( z = re^{i \theta} \), and suppose that \( \Log(z) = \log |z| + i \arg(z) \). Then for the first equation to hold, we need \[ \exp(\Log(z)) = e^{\log |z| + i \arg(z)} = z, \] that is, \( e^{\log r} = r \). The latter equation is equivalent to \( r = e^{\log r} \) for some \( k \in \mathbb{Z} \), and denoting the natural logarithm of the positive real number \( r \) by \( \ln(r) \), the former equation is equivalent to \( u = \ln |z| \). A reasonable definition of a logarithm function \( \Log \) would hence be \( \Log(z) = \ln |z| + i \arg(z) \) where \( \arg(z) \) gives the argument for the complex number \( z \) according to some convention—here is an example: ## Definition Let \( \Arg \) denote the unique argument \( \theta \) of \( z \) that lies in \( (-\pi, \pi) \) (the principal argument of \( z \)). Then the principal logarithm is the function \( \Log: \mathbb{C} \setminus \{0\} \rightarrow \mathbb{C} \) defined through \[ \Log(z) = \ln |z| + i \Arg(z). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 66 Context: Example 3.18. Here are a few evaluations of Log to illustrate this function: - Log(2) = ln(2) + Arg(2) = ln(2) - Log(-i) = ln(1) + Arg(-1) = \( \frac{3\pi}{2} \) - Log(-3) = ln(3) + Arg(-3) = ln(3) + \( \pi i \) - Log(1 - i) = ln(√2) + Arg(1 - i) = \( \frac{1}{2} \ln(2) - \frac{\pi i}{4} \) The principal logarithm is not continuous on the negative part of the real line, and so Log is a branch of the logarithm on \( \mathbb{C} \setminus \mathbb{R}^+ \). Any branch of the logarithm on a region G can be similarly extended to a function defined on \( G \setminus \{0\} \). Furthermore, the evaluation of any branch of the logarithm at a specific \( z_n \) can differ from Log(z) only by a multiple of \( 2\pi i \); the reason for this is once more the periodicity of the exponential function. So what about the second equation in (3.3), namely, Log(exp(z)) = z? Let’s try the principal logarithm: if \( z = x + iy \) then Log(exp(z)) = ln|\( z \)| + Arg(exp(z)) = ln|\( z \)| + Arg(exp(z)). The right-hand side is equal to \( z = x + iy \) if and only if \( y \in (-\pi, \pi) \). Something similar will happen with any other branch of Log of the logarithm—there will always be many z's for which Log(exp(z)) ≠ \( z \). A warning sign pointing in a similar direction concerns the much-cherished real logarithm: the ln(xy) = ln(x) + ln(y) has no analogue in \( \mathbb{C} \). For example, Log(i) + Log(-1) = \( \frac{1}{2} \ln(2 + \sqrt{2}) = \ln 2 + \frac{pi}{4} \) but Log(i - 1) = Log(-1) = \( \ln 2 - \frac{1}{2} \cdot 2 i \). The problem is that we need to come up with an argument convention to define a logarithm and then stick to this convention. There is quite a bit of subtlety here; e.g., the multi-valued map arg z := all possible arguments of z. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 67 Context: # LOGARITHMS AND COMPLEX EXPONENTIALS gives rise to a multi-valued logarithm via log z := ln |z| + i arg z. Neither arg nor log is a function, yet exp(log z) = z. We invite you to check this thoroughly; in particular, you should note how the periodicity of the exponential function takes care of the multi-valuedness of log. To end our discussion of complex logarithms on a happy note, we prove that any branch of the logarithm has the same derivative; one just has to be cautious with regions of holomorphicity. ## Proposition 3.19 If \( \text{Log}_G \) is a branch of the logarithm on G then \( \text{Log}_G \) is differentiable on G with \[ \frac{d \text{Log}_G(z)}{dz} = \frac{1}{z}. \] **Proof:** The idea is to apply Proposition 2.12 to exp and Log, but we need to be careful about the domains of these functions. Let \( H = \{ \text{Log}_G(z) : z \in G \} \), the image of Log. We apply Proposition 2.12 with \( f: H \to G \) given by \( f(z) = \text{Log}_G(z) \) and \( g: G \to H \) given by \( g(z) = \text{Log}_G(z) \); note that g is continuous, and Exercise 3.5b checks that f and g are inverses of each other. Thus Proposition 2.12 gives \[ \text{Log}_G(z) = \frac{1}{\exp(\text{Log}_G(z))} = \frac{1}{\exp(\text{Log}_G(z))} = \frac{1}{z}. \] We finish this section by defining complex exponentials. Given \( a, b \in \mathbb{C} \) with \( b \neq 0 \), the principal value of \( a^b \) is defined as \[ a^b := \exp(b \cdot \log(a)). \] Naturally, we can just as well define \( a^b \) through a different branch of the logarithm; our concern is that we use the principal value unless otherwise stated. In calculus, we can prove the equivalence of the real exponential functions as given, for \[ f(c) = e^{c}, \quad \text{where } c \in \mathbb{R}^+ \text{ and } b \text{ are defined, for example, as } c = \ln(a) + (1 + b). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 68 Context: # Examples of Functions For example, through a power series and the function \( f(z) = e^z \). With our definition of \( e^z \), we can now make a similar remark about the complex exponential function. Because \( e \) is a positive real number and hence \( \text{Arg} \, z = 0 \), \[ e^z = e^{\text{Log}(z)} = \exp\left(\log|z| + i \, \text{Arg}(z)\right) = \exp\left(\log|z|\right) = \exp(z). \] A word of caution: this only works out this nicely because we have carefully defined \( a^z \) for complex numbers. Using a different branch of logarithm in the definition of \( a^z \) can easily lead to \( e^z \neq a^z \). ## Exercises 1. Show that if \( f(z) = \frac{z^2 + 1}{z^2 - 1} \) is a Möbius transformation, then \( f^{-1}(z) = \frac{z + i}{z - i} \). 2. Complete the picture painted by Proposition 3.1 by considering Möbius transformations with \( c = 0 \). That is, show that \( f : \mathbb{C} \to \mathbb{C} \) given by \( f(z) = \frac{az + b}{cz + d} \) is a bijection, with \( f^{-1}(z) \) given by the formula in Proposition 3.1. 3. Show that (3.1) is the equation for a circle or line if and only if \( b^2 - 4ac > 0 \). Conclude that \( x + y = r \) is a solution to (3.1) if and only if \( x + iy \) is a solution to (3.2). 4. Extend Example 3.5 by showing that \( f(z) = \frac{z}{1 - z} \) maps any circle passing through \(-1\) to a line. 5. Prove that any Möbius transformation different from the identity map can have at most two fixed points. (A fixed point of a function \( f \) is a number \( z \) such that \( f(z) = z \).) 6. Prove Proposition 3.3. 7. Show that the Möbius transformation \( f(z) = \frac{az + b}{cz + d} \) maps the unit circle (minus the point \( z = 1 \)) onto the imaginary axis. 8. Suppose that \( f \) is holomorphic in the region \( G \) and \( f(G) \) is a subset of the unit circle. Show that \( f \) is constant. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 69 Context: # LOGARITHMS AND COMPLEX EXPONENTIALS **3.9** Fix \( z \in \mathbb{C} \) with \( |z| < 1 \) and consider \[ f(z) = \frac{z - a}{1 - az}. \] (a) Show that \( f(z) \) is a Möbius transformation. (b) Show that \( f^{-1}(z) = f(z) \). (c) Prove that \( f(z) \) maps the unit disk \( D(0, 1) \) to itself in a bijective fashion. --- **3.10** Suppose \[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] is a \( 2 \times 2 \) matrix of complex numbers whose determinant \( ad - bc \) is nonzero. Then we can define a corresponding Möbius transformation on \( \mathbb{C} \) by \( T_A(z) = \frac{az + b}{cz + d} \). Show that \( T_A \circ T_B = T_{AB} \), where \( \circ \) denotes composition and \( \cdot \) denotes matrix multiplication. --- **3.11** Show that our definition of \( \mathcal{C} \) honors the "finite" limit rules in Proposition 2.4, by proving the following, where \( a \in \mathbb{C} \): (a) If \( \lim_{z \to a} f(z) = \infty \) and \( \lim_{z \to b} g(z) = \infty \) then \( \lim_{z \to a} (f(z) + g(z)) = \infty \). (b) If \( \lim_{z \to a} f(z) = \infty \) and \( \lim_{z \to b} g(z) \neq 0 \) then \( \lim_{z \to a} (f(z) \cdot g(z)) = \infty \). (c) If \( \lim_{z \to a} f(z) = 0 \) and \( \lim_{z \to b} g(z) = \infty \) then \( \lim_{z \to a} (f(z) \cdot g(z)) = 0 \). (d) If \( \lim_{z \to a} f(z) = \infty \) and \( \lim_{z \to b} g(z) = \infty \) then \( \lim_{z \to a} \frac{f(z)}{g(z)} = L \) where \( L \) is finite. (e) If \( \lim_{z \to a} f(z) = 0 \) and \( \lim_{z \to b} g(z) \neq 0 \) then \( \lim_{z \to a} \frac{f(z)}{g(z)} = 0 \). --- **3.12** Let \( c_n = \frac{1}{n} \sum_{k=1}^{n} \frac{c_k}{z^k} + \dots + \frac{c_n}{z^n} \). Prove that \[ \lim_{n \to \infty} c_n = 1. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 70 Context: # EXAMPLES OF FUNCTIONS ## 3.13 Let \( f(z) = \frac{1}{z} \). Draw two graphs, one showing the following six sets in the \( x \)-plane and the other showing their images in the \( z \)-plane. Label the sets. (You should only need to calculate the images of \( \infty,\, 2,\, -1 \), and \( 1 \); remember that Möbius transformations preserve angles.) (a) the \( x \)-axis plus \( \infty \) (b) the \( y \)-axis plus \( \infty \) (c) the line \( x = y + \infty \) (d) the circle with radius 2 centered at 0 (e) the circle with radius 1 centered at 1 (f) the circle with radius 1 centered at -1 ## 3.14 Find Möbius transformations satisfying each of the following. Write your answers in standard form, as \( f(z) = \frac{az + b}{cz + d} \). (a) \( 1 \mapsto 0,\, 2 \mapsto -1,\, 3 \mapsto \infty \) (b) \( 1 \mapsto 0,\, 1 + i \mapsto 1,\, 2 \mapsto -\infty \) (c) \( 0 \mapsto -i,\, 1 \mapsto 1,\, \infty \mapsto -j \) ## 3.15 Using the cross ratio, with different choices of \( z_1 \), find two different Möbius transformations that transform \( C(1 + i) \) onto the real axis plus \( \infty \). In each case, find the image of the center of the circle. ## 3.16 Let \( y \) be the unit circle. Find a Möbius transformation that transforms \( y \) onto \( r \) and transforms \( 0 \) to \( \frac{1}{2} \). ## 3.17 Describe the image of the region under the transformation: (a) The disk \( |z| < 1 \) under \( w = \frac{z}{1 - z} \). (b) The quadrant \( x > 0,\, y > 0 \) under \( w = \frac{1}{z} \). (c) The strip \( 0 < x < 1 \) under \( w = \sqrt{z} \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 71 Context: # LOGARITHMS AND COMPLEX EXPONENTIALS ### 3.18 Find a Möbius transformation that maps the unit disk to \(\{x + iy \in \mathbb{C} : x + y > 0\}\). ### 3.19 The Jacobian of a transformation \(z = u(x, y)\), \(w = v(x, y)\) is the determinant of the matrix \[ \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix}. \] Show that if \(f = u + iv\) is holomorphic then the Jacobian equals \(|f'(z)|^2\). ### 3.20 Find the fixed points in \(C\) of \(f(z) = \frac{az + b}{cz + d}\). ### 3.21 Find each Möbius transformation \(f\): (a) \(f\) maps \(0 \mapsto 1\), \(1 \mapsto \infty\), \(\infty \mapsto 0\). (b) \(f\) maps \(1 \mapsto -1\), \(-1 \mapsto i\), and \(i \mapsto -1\). (c) \(f\) maps the \(x\)-axis to \(y = x\), the \(y\)-axis to \(y = -x\), and the unit circle to itself. ### 3.22 (a) Find a Möbius transformation that maps the unit circle to \(\{x + iy \in \mathbb{C} : x + y = 0\}\). (b) Find two Möbius transformations that map the unit disk \[ \{z \in \mathbb{C} : |z| < 1\} \] to \(\{x + iy \in \mathbb{C} : x + y > 0\}\) and \(\{x + iy \in \mathbb{C} : x + y < 0\}\), respectively. ### 3.23 Given \(a \in \mathbb{R} \setminus \{0\}\), show that the image of the line \(y = a\) under inversion is the circle with center \(\frac{1}{a}\) and radius \(\frac{1}{|a|}\). ### 3.24 Suppose \(z_1, z_2\), and \(z_3\) are distinct points in \(\mathbb{C}\). Show that \(z\) is on the circle passing through \(z_1, z_2\), and \(z_3\) if and only if \([z, z_1, z_2, z_3]\) is real or \(\infty\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 72 Context: # EXAMPLES OF FUNCTIONS ## 3.25 Prove that the stereographic projection of Proposition 3.14 is a bijection by verifying that \( \phi \circ \psi^{-1} \) and \( \psi \circ \phi^{-1} \) are the identity map. ## 3.26 Find the image of the following points under the stereographic projection \( \phi \): - \( (0, 0, -1) \) - \( (0, 0, 1) \) - \( (1, 0, 0) \) - \( (1, 1, 0) \) - \( (1, 1, 1) \) ## 3.27 Consider the plane \( H \) determined by \( x + y - z = 0 \). What is a unit normal vector to \( H \)? Compute the image of \( H \cap S^2 \) under the stereographic projection \( \phi \). ## 3.28 Prove that every circle in the extended complex plane \( \mathbb{C} \) is the image of some circle in \( S^2 \) under the stereographic projection \( \phi \). ## 3.29 Describe the effect of the basic Möbius transformations: rotation, real dilation, and translation on the Riemann sphere. *Hint:* For the first two, consider all circles in \( S^2 \) centered on the \( N \) axis, and all circles through both \( N \) and \( S \). For translation, consider two families of circles through \( N \), orthogonal to and perpendicular to the translation. ## 3.30 Show that \( \sin(i) = \sin(E) \) and \( \cos(i) = \cos(E) \). ## 3.31 Let \( z = x + iy \) and show that: (a) \( \sin z = \sin x \cosh y + i \cos x \sinh y \). (b) \( \cos z = \cos x \cosh y - i \sin x \sinh y \). ## 3.32 Prove that the zeros of \( \sin z \) are all real valued. Conclude that they are precisely the integer multiples of \( \pi \). ## 3.33 Describe the images of the following sets under the exponential function \( \exp(z) \): (a) The line segment defined by \( z = 1 + iy, \, 0 \leq y \leq 2\pi \) (b) The line segment defined by \( z = i + x, \, 0 \leq x \leq 2\pi \) (c) The rectangle defined by \( z = x + iy \in \mathbb{C}: \, 0 \leq x \leq 1, \, 0 \leq y \leq 2\pi \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 73 Context: # Logarithms and Complex Exponentials ## 3.34 Prove Proposition 3.16. ## 3.35 Prove Proposition 3.17. ## 3.36 Let \( z = x + iy \) and show that (a) \( |\sin z|^2 = \sin^2 x + \sinh^2 y = \cosh^2 y - \cos^2 x \) (b) \( |\cos z|^2 = \cos^2 x + \sinh^2 y = \cosh^2 y - \sin^2 x \) (c) If \( \cos x = 0 \) then \[ | \cot z |^2 = \frac{\cosh^2 y - 1}{\sinh^2 y} \leq 1. \] (d) If \( |y| \geq 1 \) then \[ | \cot z |^2 \leq \frac{\sinh^2 y + 1}{\sinh^2 y} = 1 + \frac{1}{\sinh^2 y} \leq 2. \] ## 3.37 Show that \( \tan(i z) = i \tanh(z) \). ## 3.38 Draw a picture of the images of vertical lines under the sine function. Do the same for the tangent function. ## 3.39 Determine the image of the strip \( \{ z \in \mathbb{C} : -\frac{\pi}{2} < \Re z < \frac{\pi}{2} \} \) under the sine function. (Hint: Exercise 3.31 makes it easy to convert parametric equations for horizontal or vertical lines to parametric equations for their images. Note that the equations \( x = A \sin t \) and \( y = B \cos t \) represent an ellipse and the equations \( x = A \cosh t \) and \( y = B \sinh t \) represent a hyperbola. Start by finding the images of the boundary lines of the strip, and then find the images of a few horizontal segments and vertical lines in the strip.) ## 3.40 Find the principal values of (a) \( \log(2) \) (b) \( (-1)^i \) (c) \( \log(-1 + i) \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 74 Context: # EXAMPLES OF FUNCTIONS ## 3.41. Convert the following expressions to the form \( z = x + iy \). (Reason carefully.) 1. \( e^z \) 2. \( e^{\cos(\log(3 + 4i))} \) 3. \( i^z \) 4. \( e^{i \pi} \) 5. \( e^{z} \) 6. \( e^{(1 + i) z} \) 7. \( \sqrt{3(1 - i)} \) 8. \( \left( \frac{1}{1} \right)^4 \) ## 3.42. Is \( \arg(E) = -\arg(z) \) true for the multiple-valued argument? What about \( \arg(E) = \Arg(z) \) for the principal branch? ## 3.43. For the multiple-valued logarithm, is there a difference between the set of all values of \( \log(z^2) \) and the set of all values of \( 2 \log(z) \)? (Hint: Try some fixed numbers for \( z \).) ## 3.44. For each of the following functions, determine all complex numbers for which the function is holomorphic. If you run into a logarithm, use the principal value unless otherwise stated. (a) \( \frac{z^2}{2} \) (b) \( \frac{1}{z^2} \) (c) \( \Log(z - 2i + 1) \) where \( \Log(z) = \ln |z| + i \arg(z) \) with \( 0 < \arg(z) < 2\pi \) (d) \( \exp(z) \) (e) \( z - 3i \) (f) \( i^z \) ## 3.45. Find all solutions to the following equations: (a) \( \Log(z) = \frac{1}{2} \) \( \cos(z) = 0 \) (b) \( \Log(z) = \frac{1}{2} \) (c) \( \exp(z) = \pi \) (d) \( \sin(z) = \cos(4) \) (e) \( \exp(z) = \exp(\sqrt{E}) \) (f) \( z = 1 + i \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 75 Context: # Logarithms and Complex Exponentials ## 3.46 Find the image of the annulus \( 1 < |z| < u \) under the principal value of the logarithm. ## 3.47 Use Exercise 2.24 to give an alternative proof that \( \log \) is holomorphic in \( \mathbb{C} \setminus \{0\} \). ## 3.48 Let \( \log_G \) be a branch of the logarithm on \( G \), and let \( H = \{ e^{g(z)} : z \in G \} \), the image of \( \log_G \). Show that \( \log_G : G \to H \) is a bijection whose inverse map is \( f(z) : H \to G \) given by \( f(z) = \exp(z) \) (i.e., \( f \) is the exponential function restricted to \( H \)). ## 3.49 Show that \( |z| = e^{\nu} \) if \( \nu \) is a positive real constant. ## 3.50 Let \( z \in \mathbb{C} \setminus \{0\} \). Find the derivative of \( f(z) = z^t \). ## 3.51 Prove that \( \exp(b \log(z)) \) is single valued if and only if \( b \) is an integer. (Note that this means that complex exponentials do not clash with monomials \( z^n \), no matter which branch of the logarithm is used.) What can you say if \( b \) is rational? ## 3.52 Describe the image under \( f \) of the line with equation \( y = x \). To do this you should find an equation (at least parametrically) for the image (you can start with the parameter form \( x = r, y = r \)), plot it reasonably carefully, and explain what happens in the limits as \( r \to -\infty \) and \( r \to \infty \). ## 3.53 For this problem, \( f(z) = z^2 \). ### (a) Show that the image under \( f \) of a circle centered at the origin is a circle centered at the origin. ### (b) Show that the image under \( f \) of a ray starting at the origin is a ray starting at the origin. ### (c) Let \( T \) be the figure formed by the horizontal segment from \( 0 \) to \( 2 \), the arc from \( 2 \) to \( 2i \), and then the vertical segment from \( 2i \) to \( 0 \). Draw \( f(T) \). ### (d) Is the right angle at the origin in part (c) preserved? Is something wrong here? (Hint: Use polar coordinates.) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 76 Context: ``` ## 3.54 As in the previous problem, let \( f(z) = z^2 \). Let \( Q \) be the square with vertices at \( 0, 2, 2+2i, \) and \( 2i \). Draw \( f(Q) \) and identify the types of image curves corresponding to the segments from \( 2 \) to \( 2+2i \) and from \( 2+2i \) to \( 2i \). They are not parts of either straight lines or circles. **Hint:** You can write the vertical segment parametrically as \( z(t) = 2 + it \). Eliminate the parameter in \( u + iv = f(t) \) to get a \( (u, v) \) equation for the image curve. Exercises 3.53 and 3.54 are related to the cover picture of this book. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 77 Context: # Chapter 4 ## Integration >If things are nice there is probably a good reason why they are nice; and if you do not know at least one reason for this good fortune, then you will have work to do. >— Richard Askey We are now ready to start integrating complex functions—and we will not stop doing so for the remainder of this book; it turns out that complex integration is much richer than real integration (in one variable). The initial reason for this is that we have an extra dimension to play with: the calculus integral \(\int_a^b f(z) dz\) has a fixed integration path, from \(a\) to \(b\) along the real line. For complex functions, there are many different ways to go from \(a\) to \(b\). ### 4.1 Definition and Basic Properties At first sight, complex integration is not really different from real integration. Let \(a, b \in \mathbb{R}\) and let \(C = [a, b]\) be continuous. Then we define \[ \int_C g(z) \, dz := \int_a^b \text{Re} \, g(t) \, dt + i \int_a^b \text{Im} \, g(t) \, dt. \tag{4.1} \] This definition is analogous to that of integration of a parametric curve in \(\mathbb{R}^2\). For a function that takes complex numbers as arguments, we typically integrate over a path \(γ\) (in place of a real interval). If you meditate about the substitution rule for real integrals (Theorem A.6), the following definition, which is based on (4.1), should come as no surprise. **Definition.** Suppose \(γ\) is a smooth path parametrized by \(γ(t)\), \(a \leq t \leq b\), and \(f\) is a complex function which is continuous on \(γ\). Then we define the integral of \(f\) on \(γ\) as \[ \int_γ f(z) \, dz = \int_a^b f(γ(t)) \, γ'(t) \, dt. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 78 Context: # Integration This definition immediately extends to paths that are piecewise smooth: Suppose \( \gamma \) is parameterized by \( \gamma(t) \), \( a \leq t \leq b \), which is smooth on the intervals \([a, c_1], [c_1, c_2], \ldots, [c_{k-1}, b]\). Then, assuming again that \( f \) is continuous on \( \gamma \), we define \[ \int_{\gamma} f \, d\gamma = \int_a^b f(\gamma(t)) \frac{d\gamma}{dt} \, dt. \] ## Example 4.1 To see this definition in action, we compute the integral of the function \( f : C \to \mathbb{R}^2 \) given by \( f(z) = z^2 \) over several paths from 0 to \( 1+i \). (a) Let \( \gamma \) be the line segment from 0 to \( 1+i \). A parameterization of this path is \( \gamma(t) = it, \; 0 \leq t \leq 1 \). Here \( \gamma'(t) = 1 \) and \( f(\gamma(t)) = (it)^2 = -t^2 \), and so: \[ \int_{\gamma} f \, d\gamma = \int_0^1 (-t^2)(1) \, dt = -\frac{1}{3}. \] (b) Let \( \beta \) be the arc of the parabola \( y = x^2 \) from 0 to \( 1+i \). A parameterization of this path is \( \beta(t) = t + it^2, \; 0 \leq t \leq 1 \). Now we have \( \beta'(t) = 1 + 2it \) and \[ f(\beta(t)) = (t + it^2)^2 = t^2 + 2it^3 - t^4, \] whence \[ \int_{\beta} f \, d\beta = \int_0^1 (t^2 + 2it^3 - t^4)(1 + 2it) \, dt = \frac{14}{3} i. \] (c) Let \( \eta \) be the union of the two line segments \( \gamma_1 \) from 0 to 1 and \( \gamma_2 \) from 1 to \( 1+i \). Parameterizations are \( \gamma_1(t) = t, \; 0 \leq t \leq 1 \) and \( \gamma_2(t) = 1 + i(t - 1), \; 0 \leq t \leq 1 \). \[ \int_{\eta} f \, d\eta = \int_{\gamma_1} f \, d\gamma + \int_{\gamma_2} f \, d\gamma = \frac{1}{3} + \frac{1}{3} i = \frac{4}{3} + \frac{2}{3} i. \] \footnotetext{1Our footnote on p. 14 about the nudity of the definition of a smooth path applies also here, at the subdivision points. Note that we do not require that the left and right derivatives match at those points.} #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 79 Context: # DEFINITION AND BASIC PROPERTIES It is apparent but nevertheless noteworthy that these integrals evaluate to different results; in particular—unlike in calculus—a complex integral does not simply depend on the endpoints of the path of integration. On the other hand, the complex integral has some standard properties, most of which follow from their real sibling in a straightforward way. Our first observation is that the actual choice of parameterization of \( \gamma \) does not matter. More precisely, if \( \gamma(t) \), \( a \leq t \leq b \) and \( \sigma(t) \), \( c \leq t \leq d \) are parameterizations of a curve, then we say that \( \sigma \) is a reparameterization of \( \gamma \) if there is an increasing piecewise smooth map of \([c, d]\) onto \([a, b]\) that takes \( t \) to \( s \), in the sense that \( \sigma = \gamma \circ r \). ## Proposition 4.2 If \( \gamma(t) \), \( a \leq t \leq b \) is a piecewise smooth parameterization of a curve and \( \sigma(t) \), \( c < s < d \) is a reparameterization of \( \gamma \), then \[ \int_{\gamma} f(\gamma(t)) \gamma'(t) \, dt = \int_{\sigma} f(\sigma(t)) \sigma'(t) \, dt. \] ## Example 4.3 To appreciate this statement, consider the two parameterizations \[ \gamma(t) = e^{it}, \quad 0 \leq t \leq 2\pi, \] and \[ \sigma(t) = 2e^{i\omega(t)}, \quad 0 \leq t \leq \frac{\pi}{2}, \] of the unit circle. Then we could write \[ \int_{\gamma} f = \int_0^{2\pi} f(\gamma(t)) \gamma'(t) \, dt \] and \[ \int_{\sigma} f = \int_0^{\frac{\pi}{2}} f(\sigma(t)) \sigma'(t) \, dt. \] A quick substitution shows that the two integrals on the respective right-hand sides are indeed equal. Proposition 4.2 says that a similar equality will hold for any integral and any parameterization. Its proof is left as Exercise 4.9, which also shows that the following definition is unchanged under reparameterization. ## Definition The length of a smooth path \( \gamma \) is \[ \text{length}(\gamma) := \int_a^b |\gamma'(t)| \, dt. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 80 Context: ``` 74 # INTEGRATION For any parametrization \( \gamma(t) \), \( a \leq t \leq b \). Naturally, the length of a piecewise smooth path \( \gamma(t) \) is the sum of the lengths of its smooth components. ## Example 4.4 Let \( \gamma \) be the line segment from \( 0 \) to \( 1 + i \), which can be parametrized by \( \gamma(t) = t + it \) for \( 0 \leq t \leq 1 \). Then: \[ \text{length}(\gamma) = \int_{0}^{1} \left| 1 + i \frac{d}{dt} \right| dt = \int_{0}^{1} \sqrt{2} dt = \sqrt{2}. \] ## Example 4.5 Let \( \gamma(t) \) be the unit circle, which can be parametrized by \( \gamma(t) = e^{it} \) for \( 0 \leq t \leq 2\pi \). Then: \[ \text{length}(\gamma) = \int_{0}^{2\pi} \left| \frac{d\gamma}{dt} \right| dt = \int_{0}^{2\pi} dt = 2\pi. \] Now we observe some basic facts about how the line integral behaves with respect to function addition, scalar multiplication, inverse parametrization, and path concatenation; we also give an upper bound for the absolute value of an integral, which we will make use of time and again. ## Proposition 4.6 Suppose \( f \) is a piecewise smooth path, \( f \) and \( g \) are complex functions which are continuous on \( \gamma \) and \( \gamma \subset \mathbb{C} \). (a) \[ \int_{\gamma} (f + g) \, dz = \int_{\gamma} f \, dz + \int_{\gamma} g \, dz. \] (b) If \( \gamma \) is parametrized by \( \gamma(t) \), \( a \leq t \leq b \), we define the path \( \gamma \) from \( \gamma(a+b) \), \( a \leq t \leq b \). Then: \[ \int_{\gamma} f \, dz = \int_{\gamma} f \, dz + \int_{\gamma} g \, dz. \] (c) If \( \gamma_1 \) and \( \gamma_2 \) are piecewise smooth paths such that \( \gamma_2 \) starts where \( \gamma_1 \) ends, we define the path \( \gamma_1 \# \gamma_2 \) by following \( \gamma_1 \) to its end and then continuing on \( \gamma_2 \) to its end. Then: \[ \int_{\gamma_1 \# \gamma_2} f \, dz = \int_{\gamma_1} f \, dz + \int_{\gamma_2} f \, dz. \] (d) \[ \left| \int_{\gamma} f \, dz \right| \leq \max |f(z)| \cdot \text{length}(\gamma). \] The path \( -\gamma \) defined in (b) is the path that we obtain by traveling through \( \gamma \) in the opposite direction. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 81 Context: # Definition and Basic Properties **Proof.** (a) follows directly from the definition of the integral and Theorem A.4, the analogous theorem from calculus. (b) follows with the real change of variables \( I = a + b - c \): \[ f = \int_{y_1}^{y_2} f(y(a + b - c)) \, dt \] \[ = -\int_{y_1}^{y_2} f(y(a + b - c)) \, dt \] \[ = \int_{y_1}^{y_2} f(y)(y(a + b - c)) \, da - \int_{y_1}^{y_2} f(y)(y) \, da = -\int f. \] (c) We need a suitable parametrization \( \gamma(t) \) for \( \gamma_1\), \( \gamma_2\). If \( \gamma_1\) has domain \([a_1, b_1]\) and \( \gamma_2\) has domain \([b_2, d_2]\), then we can use: \[ \gamma(t) = \begin{cases} \gamma_1(t) & \text{if } a_1 \leq t \leq b_1 \\ \gamma_2(t - b_1 + a_2) & \text{if } b_1 < t \leq b_2 - a_2 \end{cases} \] with domain \([a_1, b_1, b_2 - a_2]\). Now we break the integral over \( \gamma_2\) into two pieces and apply the change of variables \( I = e - b_1 + a_2\): \[ \int_{\gamma_2} f = \int_{a_1}^{b_1} f(y(t)) \, dt \] \[ = \int_{a_1}^{b_1} f(y(t)) \, dt + \int_{b_1}^{b_2} f(y(t)) \, dt. \] \[ = \int f + \int f. \] The last step follows since \( y\) restricted to \([a_1, b_1]\) is \( \gamma\) and \( y\) restricted to \([b_1 + b_2 - a_2]\) is a reparametrization of \( \gamma_2\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 82 Context: # INTEGRATION (d) Let \( \varphi = \text{Arg}(f) \). Then \( f \) is \( |f| e^{i \varphi} \) and thus, since \( |f| \in \mathbb{R} \), \[ |f| = \int_{\gamma} f = \text{Re} \left( e^{-i\varphi} \int_{\gamma} f'(t) \, dt \right) = \int_{\gamma} \text{Re}(f(t)) e^{-i\varphi(t)} \, dt \] \[ \leq \max_{t \in \gamma} |f(t)| \int_{\gamma} |f'(t)| \, |dt| \leq \max |f(z)| \cdot \text{length}(\gamma). \] Here we have used Theorem A.5 for both inequalities. ## Example 4.7 In Exercise 4.4, you are invited to show \[ \int_{\gamma} \frac{dz}{z - w} = 2\pi i, \] where \( \gamma \) is any circle centered at \( w \in \mathbb{C} \), oriented counter-clockwise. Thus Proposition 4.6(b) says that the analogous integral over a clockwise circle \(-\gamma\). Incidentally, the same example shows that the inequality in Proposition 4.6(d) is sharp: if \( \gamma \) has radius \( r \), then \[ 2\pi r = \left| \int_{\gamma} \frac{dz}{z - w} \max \frac{1}{|z - w|} \cdot \text{length}(\gamma) = \frac{1}{r} \cdot 2\pi r. \] ## 4.2 Antiderivatives The central result about integration of a real function is the Fundamental Theorem of Calculus (Theorem A.3), and our next goal is to discuss complex versions of this theorem. The Fundamental Theorem of Calculus makes a number of important claims: that continuous functions are integrable; their antiderivatives are continuous and differentiable; and that antiderivatives provide easy ways to compute values of definite integrals. The difference between the real case and the complex case is that in the latter, we need to think about integrals over arbitrary paths in \( \mathbb{C} \). ### Definition If \( F \) is holomorphic in the region \( G \subset \mathbb{C} \) and \( F'(z) = f(z) \) for all \( z \in G \), then \( F \) is an antiderivative of \( f \) on \( G \), also known as a primitive of \( f \) on \( G \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 83 Context: # ANTIDERIVATIVES ## Example 4.8 We have already seen that \( F(z) = z^2 \) is entire and has derivative \( f(z) = 2z \). Thus, \( F \) is an antiderivative of \( f \) on any region \( G \subseteq \mathbb{C} \). The same goes for \( F(z) = z^2 + C \), where \( C \in \mathbb{C} \) is any constant. ## Example 4.9 Since \[ \frac{d}{dx} \left( \frac{1}{2} ( \exp(i z) - \exp(-i z) ) \right) = \frac{1}{2} (\exp(i z) + \exp(-i z)), \] we have \[ F(z) = \sin z \] is an antiderivative of \( f(z) = \cos z \) on \( \mathbb{C} \). ## Example 4.10 The function \( F(z) = \log z \) is an antiderivative of \( f(z) = \frac{1}{z} \) on \( \mathbb{C} \setminus \{0\} \). Note that \( f \) is holomorphic in the larger region \( \mathbb{C} \setminus \{0\} \); however, we will see in Example 4.14 that \( f \) cannot have an antiderivative on that region. Here is the complex analogue of Theorem A.3(b). ## Theorem 4.11 Suppose \( G \subseteq \mathbb{C} \) is a region and \( \gamma \subseteq G \) is a piecewise smooth path with parametrization \( \gamma(t), \quad a \leq t \leq b \). If \( f \) is continuous on \( G \) and \( F \) is any antiderivative of \( f \) on \( G \) then \[ \int_{\gamma} f = F(\gamma(b)) - F(\gamma(a)). \] **Proof.** This follows immediately from the definition of a complex integral and Theorem A.3(b), since \( F'(\gamma(t)) = f(\gamma(t)) \). \[ \int_{\gamma} f = \int_a^b F'(\gamma(t)) \gamma'(t) \, dt = F(\gamma(b)) - F(\gamma(a)). \] ## Example 4.12 Since \( F(z) = \frac{1}{2} z^2 \) is an antiderivative of \( f(z) = z \) in \( \mathbb{C} \), \[ \int_{\gamma} f = \int_a^b \left( \frac{1}{2} (t^2) - \frac{1}{2} (a^2) \right) = \frac{1}{2} z^2 \] for each of the three paths in Example 4.11. There are several interesting consequences of Theorem 4.11. For starters, if \( \gamma \) is closed (that is, \( \gamma(a) = \gamma(b) \)), we effortlessly obtain the following. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 84 Context: # Integration ## Corollary 4.13 Suppose \( G \subset C \) is open, \( \gamma \) is a piecewise smooth closed path, and \( f \) is continuous on \( G \) and has an antiderivative on \( G \). Then \[ \oint_{\gamma} f = 0. \] This corollary is immediately useful as a test for existence of antiderivatives: ## Example 4.14 The function \( f : C \setminus \{ 0 \} \to C \) given by \( f(z) = \frac{1}{z} \) satisfies \( \oint_{\gamma} f(z) \, dz = 2\pi i \) for the unit circle \( \gamma \subset C \setminus \{ 0 \} \), by Exercise 4.4. Since this integral is nonzero, \( f \) cannot have an antiderivative in \( C \setminus \{ 0 \} \). We now turn to the complex analogue of Theorem A.3(a): ## Theorem 4.15 Suppose \( G \subset C \) is a region and \( z_0, z_1 \in G \). Let \( f : G \to C \) be a continuous function such that \( \oint_{\gamma} f = 0 \) for any closed piecewise smooth path \( \gamma \subset G \). Then the function \( F : G \to C \) defined by \[ F(z) = \oint_{\gamma} f, \] where \( \gamma \) is any piecewise smooth path in \( G \) from \( z_0 \) to \( z \), is an antiderivative for \( f \) on \( G \). **Proof:** There are two statements that we have to prove: first, that our definition of \( F \) is sound—that is, the integral defining \( F \) does not depend on which path we take from \( z_0 \) to \( z \)—and second, that \( F'(z) = f(z) \) for all \( z \in G \). Suppose \( G \subset C \) is a region, \( z_0, z_1 \in G \), and \( f : G \to C \) is a continuous function such that \( \oint_{\gamma} f = 0 \) for any closed piecewise smooth path \( \gamma \subset G \). Then \( f \) evaluates to the same number for any piecewise smooth path \( \gamma \) from \( z_0 \) to \( z_1 \in G \), because any two such paths \( \gamma_1 \) and \( \gamma_2 \) can be concatenated to a closed path first tracing through \( \gamma_1 \) and then through \( \gamma_2 \) backwards, which by assumption yields a zero integral: \[ \oint_{\gamma_1} f - \oint_{\gamma_2} f = 0. \] This means that \[ F(z) = \oint_{\gamma} f. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 85 Context: # Antiderivatives Let \( f \) be defined. By the same argument, \[ F(z+h) - F(z) = \int_{z}^{z+h} f = \int_{z}^{z+h} f \] for any path \( \gamma \) from \( z \) to \( z+h \). The constant function 1 has the antiderivative \( z \in C \), and so \( F = 1 \), by Theorem 4.11. Thus, \[ F(z+h) - F(z) = \frac{1}{h} \int_{z}^{z+h} (f(w) - f(z)) \, dw = \frac{1}{h} \int_{z}^{z+h} f(w) - f(z) \, dw \] If \(|h|\) is sufficiently small, then the line segment from \( z \) to \( z+h \) will be contained in \( G \), and so by applying the assumptions of our theorem for the third time, \[ F(z+h) - F(z) = \frac{1}{h} \int_{z}^{z+h} (f(w) - f(z)) \, dw = \frac{1}{h} \int_{z}^{z+h} (f(w) - f(z)) \, dw. \quad (4.2) \] We will show that the right-hand side goes to zero as \( h \to 0 \), which will conclude the theorem. Given \( \epsilon > 0 \), we can choose \( \delta > 0 \) such that \[ |w - z| < \delta \implies |f(w) - f(z)| < \epsilon \] because \( f \) is continuous at \( z \). (We can also choose \( \delta \) small enough so that (4.2) holds.) Thus, if \( |h| < \delta \), we can estimate with Proposition 6.6(1): \[ \left| \frac{1}{h} \int_{z}^{z+h} (f(w) - f(z)) \, dw \right| \leq \frac{1}{|h|} \max_{|w|<|h|} |f(w) - f(z)| \cdot \text{length}(h) = \max_{|w|<|h|} |f(w) - f(z)| < \epsilon. \] There are several variations of Theorem 4.15, as we can play with the assumptions about paths in the statement of the theorem. We give one such variation, namely, for polygonal paths, i.e., paths that are composed of unions of line segments. You should convince yourself that the proof of the following result is identical to that of Theorem 4.15. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 86 Context: Corollary 4.16. Suppose \( G \subset \mathbb{C} \) is a region and \( z_0 \in G \). Let \( f : G \to \mathbb{C} \) be a continuous function such that \[ \int_{C} f(z) \, dz = 0 \] for any closed polygonal path \( C \subset G \). Then the function \( F : G \to \mathbb{C} \) defined by \[ F(z) := \int_{\gamma_0}^{z} f, \] where \( \gamma \) is any polygonal path in \( G \) from \( z_0 \) to \( z \), is an antiderivative for \( f \) on \( G \). If you compare our proof of Theorem 4.15 to its analogue in \( \mathbb{R} \), you will see similarities, as well as some complications due to the fact that we now have to operate in the plane as opposed to the real line. Still, so far we have essentially been “doing calculus” when computing integrals. We will now take a radical departure from this philosophy by studying complex integrals that stay invariant under certain transformations of the paths we are integrating over. ### 4.3 Cauchy's Theorem The central theorem of complex analysis is based on the following concept. **Definition.** Suppose \( \gamma_0 \) and \( \gamma_1 \) are closed paths in the region \( G \subset \mathbb{C} \), parametrized by \( \gamma_0(t) \), \( 0 \leq t \leq 1 \), and \( \gamma_1(t) \), \( 0 \leq t \leq 1 \), respectively. Then \( \gamma_0 \) is \( G \)-homotopic to \( \gamma_1 \) if there exists a continuous function \( H : [0, 1] \times G \to G \) such that, for all \( t \in [0, 1] \): \[ H(0, t) = \gamma_0(t), \] \[ H(1, t) = \gamma_1(t), \] \[ H(s, 0) = H(s, 1). \] We use the notation \( \gamma_0 \sim_G \gamma_1 \) to mean \( \gamma_1 \) is \( G \)-homotopic to \( \gamma_0 \). The function \( H(t, s) \) is called a homotopy. For each fixed \( s \), a homotopy \( H(t, s) \) is a path parametrized by \( t \), and as \( s \) goes from \( 0 \) to \( 1 \), these paths continuously transform from \( \gamma_0 \) to \( \gamma_1 \). The last condition in (4.3) simply says that each of these paths is also closed. ### Example 4.17. Figure 4.1 attempts to illustrate that the unit circle is \( C(1) \), homotopic to the square with vertices \( \pm 3 \pm i \). Indeed, you should check (Exer- #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 87 Context: # Cauchy's Theorem Exercise 4.23 shows that \( \sim_0 \) is an equivalence relation on the set of closed paths in \( G \). The definition of homotopy applies to parameterization of curves; but Exercise 4.24, together with transitivity of \( \sim_0 \), shows that homotopy is invariant under reparameterizations. ## Theorem 4.18 (Cauchy’s Theorem) Suppose \( G \subset \mathbb{C} \) is a region, \( f \) is holomorphic in \( G \), \( \gamma_0 \) and \( \gamma_1 \) are piecewise smooth paths in \( G \), and \( \gamma_0 \sim \gamma_1 \). Then \[ \int_{\gamma_0} f = \int_{\gamma_1} f. \] As a historical aside, it is assumed that Johann Carl Friedrich Gauss (1777–1855) knew a version of this theorem in 1811 but published it only in 1831. Cauchy (of course) was the first to publish a complete proof. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 88 Context: # Integration Cauchy-Riemann equations fame) published his version in 1825, Karl Theodor Wilhelm Weierstrass (1815–1897) in his 1842. Theorem 4.18 is often called the **Cauchy-Goursat Theorem**, since Cauchy assumed that the derivative of \( f \) was continuous, a condition that was first removed by Édouard Jean-Baptiste Goursat (1858–1936). Before discussing the proof of Theorem 4.18, we give a basic, yet prototypical application of it: ### Example 4.19 We claim that \[ \int \frac{dz}{z} = 2 \pi i \tag{4.5} \] where \( \Gamma \) is the square in Figure 4.1, oriented counter-clockwise. We could, of course, compute this integral by hand, but it is easier to apply Cauchy’s Theorem 4.18 to the function \( f(z) = \frac{1}{z} \), which is holomorphic in \( G = \mathbb{C} \setminus \{0\} \). We showed in (4.4) that \( f \) is \( G \)-homotopic to the unit circle. Exercise 4.6 says that integrating over the unit circle gives \( 2 \pi i \) and so Cauchy’s Theorem 4.18 implies (4.5). □ ### Proof of Theorem 4.18 The full proof of Cauchy’s Theorem is beyond the scope of this book. However, there are several proofs under more restrictive hypotheses than Theorem 4.18. We shall present a proof under the following extra assumptions: 1. The derivative \( f' \) is continuous in \( G \). 2. The homotopy from \( \Gamma_0 \) to \( \Gamma_1 \) has piecewise, continuous second derivatives. Technically, this is the assumption on \( h \): \[ h(t,s) = \begin{cases} h_1(t,s) & \text{if } 0 \leq s \leq s_1, \\ h_2(t,s) & \text{if } s_1 < s \leq s_2, \\ \vdots & \\ h_k(t,s) & \text{if } s_2 < s \leq s_1, \end{cases} \] where \( h_{i}(t, s) \) has continuous second partials.* (Example 4.17 gives one instance.) Now we turn to the proof under these extra assumptions. *As we have seen with other “piecewise” definitions, the behavior of \( h \) at the subdivision lines \( s = s_i \) needs to be understood in terms of limits. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 89 Context: ```markdown # CAUCHY'S THEOREM For \( 0 \leq s \leq 1 \), let \( \gamma \) be the path parametrized by \( h(t, s) \), \( 0 \leq t \leq 1 \). Consider the function \( I : [0, 1] \to \mathbb{C} \) given by \[ I(s) := \int_{\gamma} f. \] Thus, \( I(0) = \int_{\gamma(0)} f \) and \( I(1) = \int_{\gamma(1)} f \). We will show that \( I \) is constant; in particular, \( I(0) = I(1) \), which proves the theorem. By Leibniz rule (Theorem A.9), \[ \frac{d}{ds} I(s) = \int_{\gamma(h(t,s))} f \frac{\partial h}{\partial s} \, dt = \int_{0}^{1} f\left(h(t, s)\right) \frac{\partial h(t, s)}{\partial s} \, dt \] \[ = \int_{0}^{1} \left( f\left(h(t,s)\right) \frac{\partial h(t,s)}{\partial s}\right) \, dt. \] Note that we used Theorem A.7 to switch the order of the second partials in the penultimate step—here is where we need our assumption that \( h \) has continuous second partials. Also, we needed continuity of \( f \) in order to apply Leibniz's rule. If \( b \) is piecewise defined, we split up the integral accordingly. Finally, by the Fundamental Theorem of Calculus (Theorem A.3), applied separately to the real and imaginary parts of the above integral, \[ \frac{d}{ds} I(s) = \int_{\gamma(h(t, s))} f \frac{\partial h}{\partial s} \, dt = f\left(h(t,s)\right) \bigg|_{t=0}^{t=1} = f(h(1,s)) - f(h(0,s)) \frac{\partial h}{\partial s} = 0, \] where the last step follows from \( h(0,s) = h(1,s) \) for all \( s \). **Definition.** Let \( G \subset \mathbb{C} \) be a region. If the closed path \( \gamma \) is \( G \)-homotopic to a point (that is, a constant path) then it is \( G \)-contractible, and we write \( \gamma \simeq_G \). (See Figure 4.2 for an example.) The fact that an integral over a point is zero has the following immediate consequence. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 90 Context: # Integration ![Figure 4.2](path/to/image) This ellipse is \( \mathbb{C} \setminus \mathbb{R} \)-contractible. ## Corollary 4.20 Suppose \( G \subseteq \mathbb{C} \) is a region, \( f \) is holomorphic in \( G \), \( \gamma \) is piecewise smooth, and \( \gamma \cap \partial G = \emptyset \). Then \[ \int_{\gamma} f = 0. \] This corollary is worth mediating over. For example, you should compare it with Corollary 4.13: both results give a zero integral, yet they make truly opposite assumptions (one about the existence of an antiderivative, the other about the existence of a derivative). Naturally, Corollary 4.20 gives many evaluations of integrals, such as this: ### Example 4.21 Since \( \log z \) is holomorphic in \( G = \mathbb{C} \setminus \{ 0 \} \) and the ellipse in Figure 4.2 is \( G \)-contractible, Corollary 4.20 gives \[ \int_{\gamma} \log z \, dz = 0. \] ■ ### Exercise 4.25(a) says that any closed path is \( C \)-contractible, which yields the following special case of Corollary 4.20. ## Corollary 4.22 If \( f \) is entire and \( \gamma \) is any piecewise smooth closed path, then \[ \int_{\gamma} f = 0. \] The theorems and corollaries in this section are useful not just for showing that certain integrals are zero: #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 91 Context: Example 4.23. We'd like to compute \[ \int_{C} \frac{dz}{z^2 - 2z} \] where \(C\) is the unit circle, oriented counter-clockwise. (Try computing it from first principles.) We use a partial fractions expansion to write \[ \int_{C} \frac{dz}{z^2 - 2z} = \frac{1}{2} \int_{C} \frac{dz}{z - 2} - \frac{1}{2} \int_{C} \frac{dz}{z}. \] The first integral on the right-hand side is zero by Corollary 4.20 applied to the function \(f(z) = \frac{1}{z}\) (note that \(f\) is holomorphic in \(C \setminus \{2\}\) and \(C\) is \((C \setminus \{2\})\)-contractible). The second integral is \(2\pi i\) by Exercise 4.4, and so \[ \int_{C} \frac{dz}{z^2 - 2z} = -\pi i. \] Sometimes Corollary 4.20 itself is known as Cauchy's Theorem. See Exercise 4.26 for a related formulation of Corollary 4.20, with a proof based on Green's Theorem. ### 4.4 Cauchy's Integral Formula We recall our notations: - \(C(r) = \{ z \in \mathbb{C} : |z - a| = r \}\) - \(D(r) = \{ z \in \mathbb{C} : |z - a| < r \}\) - \(D[-r] = \{ z \in \mathbb{C} : |z - a| \leq r \}\) Theorem 4.24. If \(f\) is holomorphic in an open set containing \(D[r]\) then \[ f(w) = \frac{1}{2\pi i} \int_{C(r)} f(z) \frac{dz}{z - w}. \] This is Cauchy's Integral Formula for the case that the integration path is a circle; we will prove the general statement at the end of this chapter. However, already this special case is worth meditating over: the data on the right-hand side of Theorem 4.24 #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 92 Context: # INTEGRATION Corollary 4.25. If \( f \) is holomorphic in an open set containing \( D[w, R] \), then \[ f(w) = \frac{1}{2\pi i} \int_{\mathcal{C}[w, R]} (f + R e^{i\theta}) \, d\theta, \] \[ u(w) = \frac{1}{2\pi i} \int_{0}^{2\pi} (w + R e^{i\theta}) \, d\theta \] and \[ v(w) = \frac{1}{2\pi i} \int_{0}^{2\pi} (w + R e^{i\theta}) \, d\theta. \] **Proof of Theorem 4.24 and Corollary 4.25.** By assumption, \( f \) is holomorphic in an open set \( G \) that contains \( D[w, R] \), and so \( f(z) \) is holomorphic in \( H = G \setminus \{w\} \). For any \( 0 < r < R \), \[ C[w, r] \mapsto C[w, R]. \] And so Cauchy's Theorem 4.18 and Exercise 4.4 give \[ \left| \frac{f(z)}{C[w, R]} \right|_{C[w, R]} = \frac{1}{2\pi i} \int_{C[w, R]} \frac{f(z) \, dx}{C[w, R] - z} \] \[ = \frac{1}{2\pi i} \int_{C[w, R]} \frac{f(z) - f(w)}{C[w, R] - z} \, dz \] \[ \leq S_{\max} \, \text{within} \, C[w, R] \quad (4.6) \] \[ \leq \max_{z \in C[w, R]} |f(z) - f(w)| \cdot \text{length}(C[w, r]) \cdot \frac{1}{2\pi R}. \] Here the inequality comes from Proposition 4.6(4). Now let \( \epsilon > 0 \). Because \( f \) is continuous at \( w \), there exists \( \delta > 0 \) such that \( |z - w| < \delta \) implies \[ |f(z) - f(w)| < \frac{\epsilon}{2\pi R}. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 93 Context: # CAUCHY'S INTEGRAL FORMULA In particular, this will hold for \( z \in C\setminus \{1\} \) and so (4.6) implies, with \( r = \frac{1}{2} \): \[ \left| \int_{C(r)} \frac{f(z)}{z - w} \, dz - 2\pi i f(w) \right| < \varepsilon. \] Since we can choose \( r \) as small as we like, the left-hand side must be zero, which proves Theorem 4.24. Corollary 4.25 now follows by definition of the complex integral: \[ f(w) = \frac{1}{2\pi i} \left( \int_{C(w + R\epsilon)} \frac{f(z)}{z + R\epsilon - w} \, dz = \frac{1}{2\pi i} \int_{C(w + R\epsilon)} f(z) \, dz \right), \] which splits into real and imaginary parts as \[ u(w) + i v(w) = \frac{1}{2\pi} \int_{0}^{2\pi} u(w + R e^{i\theta}) \, d\theta + i \frac{1}{2\pi} \int_{0}^{2\pi} v(w + R e^{i\theta}) \, d\theta. \] Theorem 4.24 can be used to compute integrals of a certain nature. ### Example 4.26 We'd like to determine \[ \int_{C(r)} \frac{dz}{z^2 + 1}. \] The function \( f(z) = \frac{1}{z^2 + 1} \) is holomorphic in \( C\setminus \{i, -i\} \), which contains \( \overline{D}_{1} \). Thus we can apply Theorem 4.24: \[ \int_{C(r)} \frac{dz}{z^2 + 1} = \int_{C(1)} \frac{1}{z^2 + 1} \, dz = 2\pi i f(i) = \frac{2\pi i}{2i} = \pi. \] Now we would like to extend Theorem 4.24 by replacing \( C(r) \) with any simple closed piecewise smooth path around \( z \). Intuitively, Cauchy's Theorem 4.18 should supply such an extension: assuming that \( f \) is holomorphic in a region \( G \) that includes \( z \) and is inside, we can find a small \( r \) such that \( D(w, r) \subset G \) and \[ f(w) = \frac{1}{2\pi i} \int_{C(w, r)} f(z) \, dz. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 94 Context: # Integration This all smells like good coffee, except ... we might be just dreaming. The argument may be intuitively clear, but intuition doesn’t prove anything. We'll look at it carefully, fill in the gaps, and then we'll see what we have proved. First, we need a notion of the interior of a simple closed path. The fact that any such path \( \gamma \) divides the complex plane into two connected open sets \( \gamma \) (the bounded one of which we call the inside or interior of \( \gamma \)) is one of the first substantial theorems ever proved in topology, the Jordan Curve Theorem, due to Camille Jordan (1838–1922). In this book, we shall assume the validity of the Jordan Curve Theorem. Second, we need to specify the orientation of \( \gamma \), since if the formula gives \( F(w) \) for one orientation then it will give \( -F(w) \) for the other orientation. **Definition.** A piecewise smooth closed path \( \gamma \) is positively oriented if it is parameterized so that its inside is on the left as our parameterization traverses \( \gamma \). An example is a counter-clockwise oriented circle. Third, if \( \gamma \) is positively oriented and \( D[\gamma, R] \) is a closed disk inside \( \gamma \), then we need a homotopy from \( \gamma \) to the counterclockwise circle \( C[\gamma, R] \) that stays inside \( \gamma \) and away from \( D[\gamma, R] \). This is provided directly by another substantial theorem of topology, the Arzelà–Ascoli Theorem, although there are other methods. Again, in this book we shall assume the existence of this homotopy. These results of topology seem intuitively obvious but are surprisingly difficult to prove. If you'd like to see a proof, we recommend that you take a course in topology. There is still a subtle problem with our proof. We assumed that \( \gamma \) is in \( G \), but we also need the interior of \( \gamma \) to be contained in \( G \). Since we need to apply Cauchy’s Theorem to the homotopy between \( \gamma \) and \( C[\gamma, R] \), we could just add this as an assumption to our theorem, but the following formulation will be more convenient later. **Theorem 4.27 (Cauchy’s Integral Formula).** Suppose \( f \) is holomorphic in the region \( G \) and \( \gamma \) is a positively oriented, simple, closed, piecewise smooth path, such that \( w \) is inside \( \gamma \) and \( r > 0 \). Then \[ f(w) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - w} \, dz. \] \[ \text{This is the Jordan of Jordan normal form fame, but not the one of Gauss–Jordan elimination.} \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 95 Context: So all that we need to finish the proof of Theorem 4.27 is one more fact from topology. But we can prove this one: ## Proposition 4.28 Suppose \( G \) is a simple, closed, piecewise smooth path in the region \( D \). Then \( G \) contains the interior of \( y \) if and only if \( r_g \neq 0 \). **Proof.** One direction is easy: If \( G \) contains the interior of \( y \) and \( [D, \overline{R}] \) is any closed disk in the interior of \( y \), then there is a \( G \)-homotopy from \( y \) to \( C[\overline{R}] \), and \( C[\overline{R}] \subseteq G \). In the other direction, we argue by contradiction: Assume \( r_g \cup D \) does not contain the interior of \( y \). So we can find a point \( w \) in the interior of \( y \) which is not in \( G \). Define \( g(z) = \frac{1}{z - w} \) for \( z \neq w \). Now \( g \) is holomorphic on \( G \) and \( r_g \cup D \) so Corollary 4.20 applies, and we have \( f(y, g(z)) dz = 0 \). On the other hand, choose \( R > 0 \) such that \( D[\overline{R}, R] \) is inside \( y \). There is a homotopy in \( C \) from \( y \) to \( [C, R] \), so Cauchy's Theorem 4.18, plus Exercise 4.4, shows that \( \int_{C} g(z) dz = 2\pi i \). This contradiction finishes the proof. --- Notice that, instead of using topology to prove a theorem about holomorphic functions, we just used holomorphic functions to prove a theorem about topology. ## Example 4.29 Continuing Example 4.26, Theorem 4.27 says that \[ \int_{C(2)} \frac{dz}{z^2 + 1} = \pi \] for any positively oriented, simple, closed, piecewise smooth path \( T \) that contains \( i \) on its inside and that \( C \setminus \{(C \setminus i)\} \)-contractible. ## Example 4.30 To compute \[ \int_{C(3)} \frac{\exp(z)}{z^2 - 2z} \, dz \] we use the partial fractions expansion from Example 4.23: \[ \int_{C(3)} \frac{\exp(z)}{z^2 - 2z} \, dz = \frac{1}{2} \int_{C(3)} \frac{\exp(z)}{C(z)} \, dz - \frac{1}{2} \int_{C(3)} \frac{\exp(z)}{C(z)} \, dz. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 96 Context: For the two integrals on the right-hand side, we can use Theorem 4.24 with the function \( f(z) = \exp(z) \), which is entire, and so (note that both 2 and 0 are inside \( \mathbb{C} \)): \[ \int_{C[0,3]} \frac{\exp(z)}{z^2 - 2z} \, dz = \frac{1}{2} \cdot 2\pi i \cdot \exp(2) - \frac{1}{2} \cdot 2\pi i \cdot \exp(0) = \pi (e^2 - 1). \] ## Exercises ### 4.1 Find the length of the following paths: (a) \( \gamma(t) = 3 + it, \ -1 \leq t \leq 1 \) (b) \( \gamma(t) = i + e^{it}, \ 0 \leq t \leq 1 \) (c) \( \gamma(t) = \sin(t), \ -\pi \leq t \leq \pi \) (d) \( \gamma(t) = i - e^{it}, \ 0 \leq t \leq 2\pi \) Draw pictures of each path and convince yourself that the lengths you computed are sensible. (The last path is a cycle; it is the trace of a fixed point on a wheel as it makes one rotation.) ### 4.2 Compute the lengths of the paths from Exercise 1.33: (a) the circle \( C(1 + i, 1) \) (b) the line segment from \( -1 \) to \( 2i \) (c) the top half of the circle \( C(0, 3) \) (d) the rectangle with vertices \( 1 \pm 2i \) ### 4.3 Integrate the function \( f(z) = z \) over the three paths given in Example 4.1. ### 4.4 Integrate \( \int_C f(z) \, dz \) where \( C \) is the unit circle, oriented counterclockwise. More generally, show that for any \( w \in \mathbb{C} \) and \( r > 0 \), \[ \int_{C(w,r)} \frac{dz}{z - w} = 2\pi i. \] ### 4.5 Integrate the following functions over the circle \( C(0,2) \): (a) \( f(z) = z \) (b) \( f(z) = z^2 - 2z + 3 \) (c) \( f(z) = \frac{1}{z} \) (d) \( f(z) = xy \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 97 Context: # Cauchy's Integral Formula ## 4.6 Evaluate the integrals \( \int_\gamma f(z) \, dz \), \( \int_\gamma f'(z) \, dz \), and \( \int_\gamma f''(z) \, dz \) along each of the following paths. (Hint: You can get the second two integrals after you calculate the first two by writing \( z \) as \( x + i y \).) 1. \( \gamma \) is the line segment from 0 to \( 1 - i \) 2. \( \gamma \) is \( C[0, 1] \) for some \( a \in \mathbb{C} \) ## 4.7 Evaluate \( \int_\gamma e^{\psi(z)} \, dz \) for each of the following paths: 1. \( \gamma \) is the line segment from 1 to \( i \) 2. \( \gamma \) is \( C[0, 3] \) 3. \( \gamma \) is the arc of the parabola \( y = x^2 \) from \( x = 0 \) to \( x = 1 \) ## 4.8 Compute \( \int_\gamma f(z) \, dz \) for the following functions \( f \) and paths \( \gamma \): 1. \( f(z) = z^2 \) and \( \gamma(t) = t + it^2, \, 0 \leq t \leq 1 \) 2. \( f(z) = z \) and \( \gamma \) is the semicircle from \( 1 \) through \( i \) to \( -1 \) 3. \( f(z) = e^{z} \) and \( \gamma \) is the line segment from 0 to a point \( a_0 \) 4. \( f(z) = |z|^2 \) and \( \gamma \) is the line segment from 2 to \( 3 + i \) 5. \( f(z) = z^2 + 2 + i z \) and \( \gamma \) is parameterized by \( \gamma(t), \, 0 \leq t \leq 1 \), and satisfies \( \text{Im}(\gamma(t)) > 0 \), \( \gamma(0) = 4 + i \), and \( \gamma(1) = 6 + 2i \) 6. \( f(z) = \sin(z) \) and \( \gamma \) is some piecewise smooth path from \( 0 \) to \( \pi \) ## 4.9 Prove Proposition 4.2 and the fact that the length of \( \gamma \) does not change under reparameterization. (Hint: Assume \( \gamma \) and \( \eta \) are smooth. Start with the definition of \( \int_\gamma f \), apply the chain rule to \( \sigma = \gamma \circ \eta \), and then use the change of variables formula, Theorem A.6.) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 98 Context: ``` 4.10. Prove the following integration by parts statement: Let \( f \) and \( g \) be holomorphic in \( G \), and suppose \( \gamma \subset G \) is a piecewise smooth path from \( z_0 \) to \( z_1 \). Then \[ \int_{\gamma} f(z) g'(z) \, dz = f(z_1) g(z_1) - f(z_0) g(z_0) - \int_{\gamma} f'(z) g(z) \, dz. \] 4.11. Let \( I(k) := \int_0^{2\pi} e^{ik\theta} \, d\theta \). (a) Show that \( I(0) = 1 \). (b) Show that \( I(k) = 0 \) if \( k \) is a nonzero integer. (c) What is \( I(k)^2 \)? 4.12. Compute \( \int_{C[0,1]} z^2 \, dz \). 4.13. Show that \( \int_{\gamma} z^{-1} \, dz = 0 \) for any closed piecewise smooth \( \gamma \) and any integer \( n \neq -1 \). (If \( n \) is negative, assume that \( \gamma \) does not pass through the origin, since otherwise the integral is not defined.) 4.14. Exercise 4.13 excluded \( n = -1 \) for a good reason: Exercise 4.4 gives a counterexample. Generalizing these, if \( n \) is any integer, find a closed path \( \gamma \) so that \( \int_{\gamma} z^{-1} \, dz = 2 \pi i \). 4.15. Taking the previous two exercises one step further, fix \( z_0 \in \mathbb{C} \) and let \( \gamma \) be a simple, closed, positively oriented, piecewise smooth path such that \( z_0 \) is inside \( \gamma \). Show that, for any integer \( n \), \[ \int_{\gamma} (z - z_0)^{-n} \, dz = \begin{cases} 2 \pi i & \text{if } n = -1 \\ 0 & \text{otherwise}. \end{cases} \] 4.16. Prove that \( \int_\gamma e^{z^2} \, dz = 0 \) for any closed path \( \gamma \). 4.17. Show that \( F(z) := \log(t + i) - \frac{1}{2} \log(-z - i) \) is an antiderivative of \( \frac{1}{\sqrt{z}} \) for \( \mathrm{Re}(z) > 0 \). Is \( F(z) \) equal to \( \arctan z \)? ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 99 Context: # CAUCHY'S INTEGRAL FORMULA ### 4.18. Compute the following integrals, where \( \gamma \) is the line segment from \( 4 \) to \( 4i \): 1. \( \int_{\gamma} \frac{1+z}{z} \, dz \) 2. \( \int_{\gamma} \frac{dz}{z^2 + z} \) 3. \( \int_{\gamma} \frac{z^2}{z} \, dz \) 4. \( \int_{\gamma} \sin(z^2) \, dz \) ### 4.19. Compute the following integrals. (Hint: One of these integrals is considerably easier than the other.) (a) \[ \int_{\gamma} e^{\gamma(t)} \, dt \quad \text{where } \gamma(t) = e^{it}, \quad -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \] (b) \[ \int_{\gamma} e^{\gamma(t)} \, dt \quad \text{where } \gamma(t) = e^{it}, \quad \frac{\pi}{2} \leq t \leq \frac{3\pi}{2} \] ### 4.20. Show that (4.4) gives a homotopy between the unit circle and the square with vertices \( \pm 1 \pm i \). ### 4.21. Use Exercise 1.34 to give a homotopy that proves an alternative to (4.4) and does not need a piecewise definition. ### 4.22. Suppose \( a \in \mathbb{C} \) and \( \gamma_0 \) and \( \gamma_1 \) are two counterclockwise circles so that \( a \) is inside both of them. Give a homotopy that proves \( \gamma_0 \sim_{h} \gamma_1 \). ### 4.23. Prove that \( \gamma_\infty \) is an equivalence relation. ### 4.24. Suppose that \( r \) is a closed path in a region \( G \), parameterized by \( r(t), \, t \in [0, 1] \), and it is a continuous increasing function from \( [0, 1] \) onto \( [0, 1] \). Show that \( r \) is \( G \)-homotopic to the reparameterized path \( r_0(t) = r(t) + (1-t)r \) for \( 0 \leq s \leq 1 \). ### 4.25. (a) Prove that any closed path is \( C \)-contractible. (b) Prove that any two closed paths are \( C \)-homotopic. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 100 Context: # Integration ## 4.26 This exercise gives an alternative proof of Corollary 4.20 via Green's Theorem A.10. Suppose \( G \subset \mathbb{C} \) is a region, \( f \) is holomorphic in \( G \), \( f' \) is continuous, \( \gamma \) is a simple piecewise smooth closed curve, and \( \gamma \subset G \). Explain that we may write \[ \int_\gamma f(z) \, dz = \left( \int_\gamma (x + iy) \, dx + i \int_\gamma dx - dy \right) + \int_\gamma z \, dx + i \, dy \] and show that these integrals vanish, by using Green's Theorem A.10 together with Proposition 4.28, and then the Cauchy–Riemann equations (2.2). ## 4.27 Fix \( a \in \mathbb{C} \). Compute \[ I(r) = \int_{C(r)} \frac{dz}{z - a}. \] You should get different answers for \( r < |a| \) and \( r > |a| \). (Hint: In one case \( \gamma \) is contractible in \( \mathbb{C} \setminus \{a\} \). In the other, you can combine Exercises 4.4 and 4.22.) ## 4.28 Suppose \( p(z) \) is a polynomial in \( z \) and \( \gamma \) is a closed piecewise smooth path in \( \mathbb{C} \). Show that \( \int_\gamma p = 0 \). ## 4.29 Show that \[ \int_{C[0, 2]} \frac{dz}{z^2 + 1} = 0 \] by arguing that this integral does not change if we replace \( C[0, 2] \) by \( C[0, r] \) for any \( r > 1 \), then use Proposition 4.66 to obtain an upper bound for \( |C[0, r]| \) that goes to \( 0 \) as \( r \to \infty \). ## 4.30 Compute the real integral \[ \int_0^{2\pi} \frac{d\phi}{2 + \sin \phi} \] by writing the sine function in terms of the exponential function and making the substitution \( z = e^{i\phi} \) to turn the real integral into a complex integral. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 101 Context: # Cauchy’s Integral Formula ## 4.31 Prove that for \( 0 < r < 1 \), \[ \frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{1 - r^{2}}{1 - 2r \cos(\phi) + r^{2}} \, d\phi = 1. \] (The function \( P_r(\phi) \) is the Poisson kernel and plays an important role in the world of harmonic functions, as we will see in Exercise 6.13.) ## 4.32 Suppose \( f \) and \( g \) are holomorphic in the region \( G \) and \( y \) is a simple piecewise smooth \( G \)-contractible path. Prove that if \( f(e) = g(e) \) for all \( e \in y \), then \( f(e) = g(e) \) for all \( e \) inside \( y \). ## 4.33 Show that Corollary 4.20, for simple paths, is also a corollary of Theorem 4.27. ## 4.34 Compute \[ I(r) := \int_{|z-1| = r} \frac{dz}{z^{2} + 1} \] for \( r \neq 1 \). ## 4.35 Find \[ \int_{C(0,1)} \frac{dz}{z^{2} - 2z - 8} \] for \( r = 1 \) and \( r = 5 \). (Hint: Compute a partial-fractions expansion of the integrand.) ## 4.36 Use the Cauchy Integral Formula (Theorem 4.24) to evaluate the integral in Exercise 4.35 when \( r = 3 \). ## 4.37 Compute the following integrals. (a) \[ \int_{C(-1, 3)} \frac{z^{2}}{z^{2} - 4} \, dz \] (b) \[ \int_{C(1)} \sin z \, dz \] (c) \[ \int_{C(2)} \frac{\exp(z)}{z^{3}} \, dz \] (d) \[ \int_{C(2)} \frac{\exp(z)}{z^{4}} \, dz \] *Named after Siméon Denis Poisson (1781–1840).* #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 102 Context: ``` 4.38 Let \( f(z) = \frac{1}{z} \) and define the two paths \( \gamma = C_{1.1} \) oriented counterclockwise and \( a = C_{1,-1} \) oriented clockwise. Show that \( f \circ \gamma = \int_{\gamma} f \, dz \) even though \( \gamma \) goes through \( G = \mathbb{C} \setminus \{ 0 \} \), the region of holomorphicity of \( f \). 4.39 This exercise gives an alternative proof of Cauchy's Integral Formula (Theorem 4.27) that does not depend on Cauchy's Theorem (Theorem 4.18). Suppose the region \( G \) is convex; this means that, whenever \( u \) and \( w \) are in \( G \), the line segment between them is also in \( G \). Suppose \( f \) is holomorphic in \( G \), \( f' \) is continuous, and \( \gamma \) is a positively oriented, simple, closed, piecewise smooth path, such that \( w \) is inside \( \gamma \) and \( \gamma \) is simple. (a) Consider the function \( g: [0,1] \to \mathbb{C} \) given by \[ g(t) = \int_{\gamma} \frac{f(u + t(u - w))}{z - w} \, dz. \] Show that \( g'(t) = 0 \). (Hint: Use Theorem A.9 (Leibniz's rule) and then find an antiderivative for \( \frac{f(u + t(u - w))}{z - w} \).) (b) Prove Theorem 4.27 by evaluating \( g(0) \) and \( g(1) \). (c) Why did we assume \( G \) is convex? ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 103 Context: # Chapter 5 ## Consequences of Cauchy’s Theorem > Everybody knows that mathematics is about miracles; only mathematicians have a name for these miracles. > — Roger Howe Cauchy’s Theorem and Integral Formula (Theorems 4.18 and 4.27), which we now have at our fingertips, are not just beautiful results but also incredibly practical. In a quite concrete sense, the rest of this book will reap the fruits that these two theorems provide us with. This chapter starts with a few highlights. ### 5.1 Variations of a Theme We now derive formulas for \( f' \) and \( f'' \) which resemble Cauchy’s Integral Formula (Theorem 4.27). **Theorem 5.1** Suppose \( f \) is holomorphic in the region \( G \) and \( \gamma \) is a positively oriented, simple, closed, piecewise smooth, \( G \)-contractible path. If \( w \) is inside \( \gamma \) then \[ f'(w) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{(z - w)^2} \, dz. \] Moreover, \( f''(w) \) exists, and \[ f''(w) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{(z - w)^3} \, dz. \] **Proof:** The idea of our proof is very similar to that of Cauchy’s Integral Formula (Theorems 4.24 and 4.27). We will study the following difference quotient, which... #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 104 Context: we rewrite using Theorem 4.27. $$ \frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{1}{2\pi i} \int_{\gamma} \frac{f(t)}{t - z} \, dt - \frac{1}{2\pi i} \int_{\gamma} \frac{f(t)}{(t - z)^2} \, dz $$ $$ = \frac{1}{2\pi i} \int_{\gamma} \frac{f(t)}{(t - z)^2} \, dt. \tag{5.1} $$ Theorem 5.1 will follow if we can show that the following expression gets arbitrarily small as \( \Delta z \to 0 \): $$ \frac{f(z + \Delta z) - f(z)}{\Delta z} = \frac{1}{2\pi i} \int_{\gamma} \frac{f(t)}{(t - z)^2} \, dt - \frac{f(z)}{(z - w)^2} \, dz $$ This can be made arbitrarily small if we can show that the integral on the right-hand side stays bounded as \( \Delta z \to 0 \). In fact, by Proposition 4.6, it suffices to show that the integrand stays bounded as \( \Delta z \to 0 \) (because \( z \) and curve length \( \gamma \) are fixed). Let \( M = \max_{t \in \gamma} |f(t)| \) (whose existence is guaranteed by Theorem A1). Choose \( \delta > 0 \) such that \( |z - w| < \delta \): that is, \( |z - w| \geq \epsilon \) for all \( y \in \gamma \). By the reverse triangle inequality (Corollary 1.7(b)), for all \( y \in \gamma \): $$ \frac{|f(t)|}{|t - z|^2} \leq \frac{|f(z)|}{|z - w|^2} \leq \frac{M}{(a - |z - w|)^2}, $$ which certainly stays bounded as \( \Delta z \to 0 \). This proves (5.1) and thus the Cauchy Integral Formula for \( f' \). The proof of the formula for \( f'' \) is very similar and will be left to Exercise 5.2. Theorem 5.1 suggests that there are similar formulas for the higher derivatives of \( f \). This is in fact true, and theoretically we could obtain them one by one with the methods of the proof of Theorem 5.1. However, once we start studying power series for holomorphic functions, we will obtain such a result much more easily; so we save the derivation of integral formulas for higher derivatives of \( f \) for later (Corollary 8.11). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 105 Context: # VARIATIONS OF A THEME Theorem 5.1 has several important consequences. For starters, it can be used to compute certain integrals. ## Example 5.2 \[ \int_{C(0,1)} \frac{\sin(z)}{z^2} \, dz = \frac{2\pi i}{2\pi} \left. \frac{d}{dx} \sin(z) \right|_{z=0} = 2\pi i \cos(0) = 2\pi i. \] ![Figure 5.1: The integration paths in Example 5.3.](path/to/figure5.1.png) ## Example 5.3 To compute the integral \[ \int_{C(0,2)} \frac{dz}{z^2(z-1)}, \] we could employ a partial fractions expansion similar to the one in Example 4.23, or moving the integration path similar to the one in Exercise 4.29. To exhibit an alternative, we split up the integration path as illustrated in Figure 5.1: we introduce an additional path that separates 0 and 1. If we integrate on these two new closed paths (\(Y_1\) and \(Y_2\)) counterclockwise, the two contributions along the new path will cancel each other. The effect is that we transformed an integral for which two singularities were inside the integration path into a sum of two integrals, each of which has only one singularity inside the integration path; these new integrals will... #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 106 Context: # Consequences of Cauchy’s Theorem We know how to deal with, using Theorems 4.24 and 5.1: \[ \int_{\mathcal{C}(0)} \frac{dz}{z^2(z-1)} = \int_{\mathcal{C}(1)} \frac{dz}{z(z-1)^2} = 2\pi i \left( \frac{1}{z} \bigg|_{z=1} + \frac{1}{(-1)^2} \right) = 2\pi i \left( 1 + \frac{1}{2} \right) = 0. \] ## Example 5.4 \[ \int_{\mathcal{C}(1)} \frac{\cos(z)}{z^3} dz = \int_{\mathcal{C}(1)} \frac{d^3}{dz^3} \cos(z) \bigg|_{z=0} = \pi i(-\cos(0)) = -\pi i. \] Theorem 5.1 has another powerful consequence; just from knowing that \( f \) is holomorphic in \( G \), we know of the existence of \( f'' \), that \( f' \) is also holomorphic in \( G \). Repeating this argument for \( f' \), \( f'' \), etc., shows that all derivatives \( f^{(n)} \) exist and are holomorphic. We can translate this into the language of partial derivatives; since the Cauchy–Riemann equations (Theorem 2.13) show that any sequence of partial differentiations of \( f \) results in a constant times \( f^{(n)} \). ## Corollary 5.5 If \( f \) is differentiable in a region \( G \) then \( f \) is infinitely differentiable in \( G \), and all partials of \( f \) with respect to \( x \) and \( y \) exist and are continuous. ## 5.2 Antiderivatives Again Theorem 4.15 gives us an antiderivative for a function that has zero integrals over closed paths in a given region. Now that we have Corollary 5.5, mediating just a bit more over Theorem 4.15 gives a converse of sorts to Corollary 4.20. ### Corollary 5.6 (Morera's Theorem) Suppose \( f \) is continuous in the region \( G \) and \[ \int_{C} f = 0 \] for all piecewise smooth closed paths \( C \subset G \). Then \( f \) is holomorphic in \( G \). [^1]: Named after Giacinto Morera (1856–1907). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 107 Context: Proof: Theorem 4.15 yields an antiderivative \( F \) for \( f \) in \( G \). Because \( F \) is holomorphic in \( G \), Corollary 5.5 implies that \( f \) is also holomorphic in \( G \). Just like there are several variations of Theorem 4.15, we have variations of Corollary 5.6. For example, by Corollary 4.16, we can replace the condition for all piecewise smooth closed paths \( \gamma \) in the statement of Corollary 5.6 by the condition for every closed path in \( G \) (which, in fact, gives a stronger version of this result). A special case of Theorem 4.15 applies to regions in which every closed path is contractible. ### Definition A region \( G \subset \mathbb{C} \) is **simply connected** if \( \gamma \) is contractible for every closed path \( \gamma \) in \( G \). ### Example 5.7 Any disk \( D_r \) is simply connected, as is \( \mathbb{C} \setminus \{0\} \). (You should draw a few closed paths in \( \mathbb{C} \setminus \{0\} \) to convince yourself that they are all contractible.) The region \( \{0\} \) is not simply connected, as e.g., the unit circle is not \( \{(0)\} \)-contractible. If \( f \) is holomorphic in a simply-connected region then Corollary 4.20 implies that \( f \) satisfies the conditions of Theorem 4.15, hence we conclude: ### Corollary 5.8 Every holomorphic function on a simply-connected region \( G \subset \mathbb{C} \) has an antiderivative on \( G \). Note that this corollary gives us no indication of how to compute an antiderivative. For example, it says that the (entire) function \( f : \mathbb{C} \to \mathbb{C} \) given by \( f(z) = \exp(z^2) \) has an antiderivative \( F \) in \( C \); it is an entirely different matter to derive a formula for \( F \). Corollary 5.8 also illustrates the role played by two of the regions in Example 5.7, in connection with the function \( f(z) = \frac{1}{z} \). This function has no antiderivative on \( \mathbb{C} \setminus \{0\} \), as was proved in Example 4.14. Consequently, since \( \mathbb{C} \setminus \{0\} \) is not simply connected, however, the function \( f(z) = \frac{1}{z} \) does not illustrate the instance implied by Corollary 5.8. Finally, Corollary 5.8 implies that, if we have two paths in a simply-connected region with the same endpoints, we can concatenate them—changing direction on one—to form a closed path, which proves: #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 108 Context: # Consequences of Cauchy’s Theorem ## Corollary 5.9 If \( f \) is holomorphic in a simply-connected region \( G \) then \( f \) is independent of the piecewise smooth path \( \gamma \) between \( y(a) \) and \( y(b) \). When an integral depends only on the endpoints of the path, the integral is called path independent. Example 41 shows that this situation is quite special; it also says that the function \( z^2 \) does not have an antiderivative in, for example, the region \( \{ z \in \mathbb{C} : |z| < 2 \} \). (Actually, the function \( z^2 \) does not have an antiderivative in any nonempty region — see Exercise 5.7.) ## 5.3 Taking Cauchy’s Formula to the Limit Many beautiful applications of Cauchy’s Integral Formulas (such as Theorems 4.27 and 5.1) arise from considerations of the limiting behavior of the integral as the path gets arbitrarily large. The first and most famous application concerns the roots of polynomials. As a preparation we prove the following inequality, which is generally quite useful. It says that for \( | z | \) large enough, a polynomial \( p(z) \) of degree \( a_n \) tends to behave almost like a constant times \( z^{a_n} \). ### Proposition 5.10 Suppose \( p(z) \) is a polynomial of degree \( a_n \) with leading coefficient \( a_n \). Then there is a real number \( R \) such that \[ \frac{1}{2} | z |^{a_n} \leq | p(z) | \leq 2 | a_n | | z |^{a_n} \] for all \( z \) satisfying \( | z | \geq R \). **Proof:** Since \( p(z) \) has degree \( a_n \), its leading coefficient \( a_n \) is not zero, and we can factor out \( a_n z^{a_n} \): \[ | p(z) | = | a_n z^{a_n} + a_{n-1} z^{a_n-1} + \ldots + a_0 | = | a_n | \cdot \left| z^{a_n} \left(1 + \frac{a_{n-1}}{a_n} \frac{1}{z} + \ldots + \frac{a_0}{a_n} \frac{1}{z^{a_n}} \right) \right|. \] Then the term inside the last factor has limit as \( z \to \infty \) (by Exercise 3.12), and so its modulus is between \( \frac{1}{2} \) and \( 2 \) as long as \( | z | \) is large enough. ## Theorem 5.11 (Fundamental Theorem of Algebra) Every nonconstant polynomial has a root in \( \mathbb{C} \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 109 Context: ```markdown # Taking Cauchy’s Formulas to the Limit ## Proof Suppose (by way of contradiction) that \( p(z) \) does not have any roots, that is, \( p(z) \neq 0 \) for all \( z \in \mathbb{C} \). Then \( \frac{1}{p(z)} \) is entire, and so Cauchy’s Integral Formula (Theorem 4.24) gives \[ \frac{1}{p(0)} = \frac{1}{2\pi i} \int_{\mathcal{C}(0, R)} \frac{p'(z)}{p(z)} \, dz, \quad \text{for any } R > 0. \] Let \( d \) be the degree of \( p(z) \) and \( a_k \) its leading coefficient. Propositions 4.6(d) and 5.10 allow us to estimate, for sufficiently large \( R \): \[ \left| \frac{1}{p(0)} \right| = \frac{1}{2\pi} \int_{\mathcal{C}(0, R)} \frac{1}{|p(z)|} \, |dz| \leq \frac{1}{2\pi} \cdot \frac{1}{\text{ext } |a_k| R^{d}} \cdot |2\pi R| = \frac{2 |a_k|}{R^{d}}. \] The left-hand side is independent of \( R \), while the right-hand side can be made arbitrarily small (by choosing \( R \) sufficiently large), and so we conclude that \( \frac{1}{p(0)} = 0 \), which is impossible. ## Theorem 5.11 Theorem 5.11 implies that any polynomial \( p \) can be factored into linear terms of the form \( z - a \) where \( a \) is a root of \( p \) as we can apply the corollary, after getting a root to \( z^n = 0 \) (which is again a polynomial by the division algorithm), etc. (see also Exercise 5.11). A compact reformulation of the Fundamental Theorem of Algebra (Theorem 5.11) is to say that \( \mathbb{C} \) is algebraically closed in contrast, \( \mathbb{R} \) is not algebraically closed. ### Example 5.12 The polynomial \( p(z) = z^4 + 5z^2 + 3 \) has no roots in \( \mathbb{R} \). The Fundamental Theorem of Algebra (Theorem 5.11) states that \( p \) must have a root (in fact, four roots) in \( \mathbb{C} \): \[ p(z) = (z^2 + 1)(z^2 + 3) = (z + i)(z - i)(\sqrt{3} + z)(\sqrt{3} - z) \quad \Box \] Another powerful consequence of Theorem 5.1 is the following result, which again has no counterpart in real analysis (consider, for example, the ratio test). ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 110 Context: # Corollary 5.13 (Liouville's Theorem) Any bounded entire function is constant. ## Proof: Suppose \( |f(z)| \leq M \) for all \( z \in \mathbb{C} \). Given any \( \epsilon > 0 \), we apply Theorem 5.1 with the circle \( C(R) \); note that we can choose any \( R > 0 \) because \( f \) is entire. By Proposition 4.16(1), \[ \left| \frac{f(w)}{2\pi i \int_{C(R)} (z - w)^{-1} dz} \right| = \frac{1}{2\pi i} \int_{C(R)} \frac{f(z)}{(z - w)^2} dz \leq \frac{1}{2\pi} \max_{z \in C(R)} |f(z)| \cdot \frac{1}{2\pi R}. \] The right-hand side can be made arbitrarily small, as we are allowed to choose \( R \) as large as we want. This implies that \( f' = 0 \), and hence, by Theorem 2.17, \( f \) is constant. As an example of the usefulness of Liouville's theorem (Corollary 5.13), we give another proof of the Fundamental Theorem of Algebra, close to Gauss's original proof. ## Second proof of Theorem 5.11 (Fundamental Theorem of Algebra): Suppose (by way of contradiction) that \( f \) does not have any roots, that is, \( f(z) \neq 0 \) for all \( z \in \mathbb{C} \). Then the function \( f(z) = \frac{1}{f(z)} \) is entire. But \( f \to 0 \) as \( |z| \to \infty \), by Proposition 5.10; consequently, by Exercise 5.10, \( f \) is bounded. Now we apply Corollary 5.13 to deduce that \( f \) is constant. Hence \( f \) is constant, which contradicts our assumptions. As one more example of the theme of getting results from Cauchy's Integral Formulas by taking the limit as a path “goes to infinity,” we compute an improper integral. ## Example 5.14: We will compute the (real) integral \[ \int_0^{\infty} \frac{dx}{x^2 + 1} = \pi. \] Let \( C_R \) be the counterclockwise semicircle formed by the segment \([-R, R]\) of the real axis from \(-R\) to \(R\), followed by the circular arc \( \gamma_R \) of radius \( R \) in the upper half plane from \( R \to -R \), see Figure 5.2. > *This theorem (the historical reason commonly attributed to Joseph Liouville (1809–1883). It was published earlier by Cauchy in fact, Gauss may well have known about it before Cauchy.)* #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 111 Context: # TAKING CAUCHY'S FORMULAS TO THE LIMIT ## Figure 5.2: The integration paths in Example 5.14 We computed the integral over \( \sigma_R \) already in Example 4.29: \[ \int_{-R}^{R} \frac{dz}{z^2 + 1} = \pi. \] This holds for any \( R > 1 \), and so we can take the limit as \( R \to \infty \). By Proposition 4.6(d) and the reverse triangle inequality (Corollary 1.7(b)): \[ \left| \int_{-R}^{R} \frac{dz}{z^2 + 1} \right| \leq \max_{\sigma_R} \left| \frac{1}{z^2 + 1} \right| \cdot |R| \leq \max_{\sigma_R} \left( \frac{1}{|z|^2 - 1} \right) R = \frac{R}{R^2 - 1} \] which goes to \( 0 \) as \( R \to \infty \). Thus, \[ \pi = \lim_{R \to \infty} \int_{-R}^{R} \frac{dz}{z^2 + 1} = \lim_{R \to \infty} \left( \int_{-\infty}^{-R} \frac{dz}{z^2 + 1} + \int_{R}^{\infty} \frac{dz}{z^2 + 1} \right). \] Of course, this integral can be evaluated almost as easily using standard formulas from calculus. However, just slight modifications of this example lead to improper integrals that are beyond the scope of basic calculus; see Exercises 5.18 and 5.19. ########## """QUERY: Summarize in 5 sentences the contents of your context. Speak in Old English manner similar to how Shakespeare writes his sonnets.""" Consider the chat history for relevant information. Use all information included. Use as much tokens as needed. Important: If you find information separated by a | in the context, it is a table formatted in Markdown.Important: Take a look at the QUERY and only the QUERY. If this is vague or unclear, please ignore everything and ask a follow-up question instead! 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Page 102, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 103, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 104, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 105, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 106, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 107, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 108, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 109, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 110, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 111 ================================================== **Elapsed Time: 7.44 seconds** ================================================== FINAL ANSWER Answer: In yon document, a report doth unfold, wherein Lois Layne Ramos, a diligent tester, hath scrutinized a web application with keen eyes. She hath discovered several flaws, such as the input field's acceptance of empty values and the perilous absence of a login page, which doth expose sensitive data to the whims of the uninvited. The report, rich in detail, doth provide steps to replicate each issue, alongside images that illustrate the user interface, thus guiding developers to mend the faults. Furthermore, it doth recommend enhancements to the application, ensuring a more secure and user-friendly experience for all who venture therein. Thus, the findings serve as a beacon, illuminating the path toward a more robust and trustworthy application, worthy of its users' trust. ================================================== **Elapsed Time: 4.16 seconds** ==================================================