{ "query": "Please summarize the whole context. It is important that you include a summary for each file. All files should be included, so please make sure to go through the entire context", "namespace": "cdf5e841-bd92-4996-b7f2-7c2a398184c5", "messages": [], "stream": false, "language_level": "", "chat_channel": "", "language": "German", "tone": "neutral", "writing_style": "standard", "model": "gemini-1.5-flash", "knowledgebase": "ki-dev-large", "seed": 0, "client_id": 0, "all_context": true, "follow_up_for": null, "knowledgebase_files_count": 0, "override_command": "", "disable_clarity_check": true, "custom_primer": "", "logging": true, "query_route": "" } INITIALIZATION Knowledgebase: ki-dev-large Base Query: Please summarize the whole context. It is important that you include a summary for each file. All files should be included, so please make sure to go through the entire context Model: gemini-1.5-flash Use Curl?: ================================================== **Elapsed Time: 0.00 seconds** ================================================== ROUTING Query type: summary ================================================== **Elapsed Time: 2.18 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: 0.39 seconds** ================================================== FINAL QUERY Final Query: CONTEXT: ########## File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 1 Context: # A First Course in Complex Analysis **Version 1.54** ## Authors - **Matthias Beck** Department of Mathematics San Francisco State University San Francisco, CA 94132 matthec@sfus.edu - **Gerald Marchesi** Department of Mathematical Sciences Binghamton University (SUNY) Binghamton, NY 13902 marchesi@math.binghamton.edu - **Dennis Pixton** Department of Mathematical Sciences Binghamton University (SUNY) Binghamton, NY 13902 dennis@math.binghamton.edu - **Lucas Sabalka** Lincoln, NE 68502 sabalka@ergs1.com #################### 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 112 Context: # Exercises ## 5.1 Compute the following integrals, where \(\square\) is the boundary of the square with vertices at \(\pm 4i\), positively oriented: (a) \(\int_{\square} \frac{e^{z^2}}{z^2} \, dz\) (b) \(\int_{\square} \frac{z^3}{(x - \pi^2)} \, dz\) (c) \(\int_{\square} \frac{\sin(2z)}{(z - \pi)^2} \, dz\) (d) \(\int_{\square} \frac{e^{z} \cos(z)}{(x - \pi)} \, dz\) ## 5.2 Prove the formula for \(f^{(n)}\) in Theorem 5.1. *Hint: Modify the proof of the integral formula for \(f^{(n)}(w)\) as follows:* (a) Write a difference quotient for \(f^{(n)}(w)\), and use the formula for \(f^{(n)}(w)\) in Theorem 5.1 to convert this difference quotient into an integral of \(f(z)\) divided by some polynomial. (b) Subtract the desired integral formula for \(f^{(n)}\) from your integral for the difference quotient, and simplify to get the analogue of (5.1). (c) Find a bound as in the proof of Theorem 5.1 for the integrand, and conclude that the limit of the difference quotient is the desired integral formula. ## 5.3 Integrate the following functions over the circle \(C(0,3)\): (a) \(\log(z - 4i)\) (b) \(\frac{1}{z - \frac{1}{2}}\) (c) \(\frac{1}{z^2}\) (d) \(\frac{e^{z}}{z^3}\) (e) \(\cos(z) \, z\) (f) \(z^{-3}\) (g) \(\frac{\sin(z)}{(z^2 + 1)}\) (h) \(\frac{1}{(4 + z^2 + 1)}\) (i) \(\frac{e^{2z}}{(z - 1)(\sqrt{z - 2})}\) ## 5.4 Compute \(\int_{C(2)} \frac{e^{z}}{(z - w)^2}\, dz\) where \(w\) is any fixed complex number with \(|w| \neq 2\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 113 Context: 5.5. Define \( f : [0, 1] \to \mathbb{C} \) through \[ f(x) := \int_0^1 \frac{dw}{1 - wx} \] (the integration path is from 0 to 1 along the real line). Prove that \( f \) is holomorphic in the unit disk \( D(0, 1) \). 5.6. To appreciate Corollary 5.5, show that the function \( f : \mathbb{R} \to \mathbb{R} \) given by \[ f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \] is differentiable in \( \mathbb{R} \), yet \( f' \) is not even continuous (much less differentiable) at 0. 5.7. Prove that \( f(2) = e^2 \) does not have an antiderivative in any nonempty region. 5.8. Show that \( \exp(\sin z) \) has an antiderivative on \( \mathbb{C} \). (What is it?) 5.9. Find a region on which \( f(z) = \exp(z^2) \) has an antiderivative. (Your region should be as large as you can make it. How does this compare with the real function \( f(x) = e^x \)?) 5.10. Suppose \( f \) is continuous on \( \mathbb{C} \) and \( \lim_{z \to \infty} f(z) = L \). Show that \( f \) is bounded. (Hint: If \( \lim_{z \to \infty} f(z) = L \), use the definition of the limit at infinity to show that there is \( R > 0 \) so that \( |f(z) - L| < \epsilon \) if \( |z| > R \). Now argue that \( |f(z)| < |L| + 1 \) for \( |z| \geq R \). Use an argument from calculus to show that \( |f(z)| \) is bounded for \( |z| \leq R \).) 5.11. Let \( p(z) \) be a polynomial of degree \( n > 0 \). Prove that there exist complex numbers \( c_1, c_2, \ldots, c_n \), and positive integers \( j_1, \ldots, j_n \) such that \[ p(z) = c e^{-j_1} (z - z_1)^{j_1} \cdots (z - z_n)^{j_n} \] where \( j_1 + \ldots + j_n = n \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 114 Context: 5.12. Show that a polynomial of odd degree with real coefficients must have a real zero. *(Hint: Use Exercise 1.24.)* 5.13. Suppose \( f \) is entire and \( |f(z)| \leq \sqrt{|z|} \) for all \( z \in \mathbb{C} \). Prove that \( f \) is identically 0. *(Hint: Show first that \( f \) is constant.)* 5.14. Suppose \( f \) is entire and there exists \( M > 0 \) such that \( |f(z)| \leq M \) for all \( z \in \mathbb{C} \). Prove that \( f \) is constant. 5.15. Suppose \( f \) is entire with bounded real part, i.e., writing \( f(z) = u(z) + iv(z) \), there exists \( M > 0 \) such that \( |u(z)| \leq M \) for all \( z \in \mathbb{C} \). Prove that \( f \) is constant. *(Hint: Consider the function \( \exp(f(z)) \).)* 5.16. Suppose \( f \) is entire and there exist constants \( a \) and \( b \) such that \( |f(z)| \leq |z|^a + b \) for all \( z \in \mathbb{C} \). Prove that \( f \) is polynomial of degree at most 1. *(Hint: Use Theorem 5.1 and Exercise 2.29.)* 5.17. Suppose \( f: D(0, 1) \to D(0, 1) \) is holomorphic. Prove that for \( |z| < 1 \), \[ |f'(z)| \leq \frac{1}{1 - |z|}. \] 5.18. Compute \[ \int_0^{\infty} \frac{dx}{x^4 + 1}. \] 5.19. In this problem \( f(z) = \frac{e^{iz}}{e^{iz} + 1} \) and \( R > 1 \). Modify our computations in Example 5.14 as follows: (a) Show that \( f^* = f \) where \( f^* \) is again (as in Figure 5.2) the counterclockwise semicircle formed by the segment \([-R, R]\) on the real axis, followed by the circular arc \( \gamma_R \) of radius \( R \) in the upper half plane from \( R \) to \(-R\). (b) Show that \( |f(z)| \leq \frac{1}{\sqrt{2}} \) for \( z \) in the upper half plane, and conclude that \( \lim_{|z| \to \infty} f(z) = 0 \) and hence \( \lim_{R \to \infty} f_{R} = 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 115 Context: # Taking Cauchy's Formulas to the Limit ## 5.20. Compute \[ \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+1} \, dx. \] ## 5.21. This exercise outlines how to extend some of the results of this chapter to the Riemann sphere as defined in Section 3.2. Suppose \( G \subset \mathbb{C} \) is a region that contains 0, let \( f \) be a continuous function on \( G \), and let \( \gamma \subset G \) be a piecewise smooth path in \( G \) avoiding the origin, parametrized as \( \gamma(t) \), for \( a \leq t \leq b \). ### (a) Show that \[ \int_{\gamma} f(z) \, dz = \int_{a}^{b} \frac{f(\gamma(t))}{\gamma'(t)} \, dt \] where \( o(t) = \gamma(t) \), \( a \leq t \leq b \). Now suppose \(\mathrm{Im}(f) = \{ f(z) \} \) is finite. Let \( H = \{ z \in G \setminus \{ 0 \} \} \) and define the function \( g : H \cup \{ 0 \} \to \mathbb{C} \) by \[ g(z) = \begin{cases} \frac{f(z)}{L} & \text{if } z \in H, \\ 0 & \text{if } z = 0. \end{cases} \] Thus \( g \) is continuous on \( H \cup \{ 0 \} \) and gives the identity \[ \int_{H} f \, dz = \int g. \] In particular, we can transfer certain properties between these two integrals. For example, if \( f \) is path independent, so is \( g \). Here is but one application: ### (a) Show that \[ \int_{\gamma} f(z) \, dz \text{ is path independent for any integer } n \neq -1. \] ### (b) Conclude (once more) that \[ \int z^n \, dz = 0 \text{ for any integer } n \neq -1. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 116 Context: # Chapter 6 ## Harmonic Functions The shortest route between two truths in the real domain passes through the complex domain. Jacques Hadamard (1865–1963) We will now spend a short while on certain functions defined on subsets of the complex plane that are real valued, namely those functions that are harmonic in some region. The main motivation for studying harmonic functions is that the partial differential equation they satisfy is very common in the physical sciences. Their definition briefly showed its face in Chapter 2, but we study them only now in more detail, since we have more machinery at our disposal. This machinery comes from complex-valued functions, which are, nevertheless, intimately connected to harmonic functions. ### 6.1 Definition and Basic Properties Recall from Section 2.3 the definition of a harmonic function: **Definition.** Let \( G \subset \mathbb{C} \) be a region. A function \( u : G \rightarrow \mathbb{R} \) is harmonic in \( G \) if it has continuous second partials in \( G \) and satisfies the Laplace* equation \[ u_{xx} + u_{yy} = 0. \] **Example 6.1.** The function \( u(x, y) = xy \) is harmonic in \( C \) since \( u_{xx} + u_{yy} = 0 + 0 = 0. \) **Example 6.2.** The function \( u(x, y) = e^x \cos(y) \) is harmonic in \( C \) because \[ u_{xx} + u_{yy} = e^x \cos(y) - e^x \cos(y) = 0. \] 1. Named after Pierre-Simon Laplace (1749–1827). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 117 Context: There are (at least) two reasons why harmonic functions are part of the study of complex analysis, and they can be found in the next two theorems. ## Proposition 6.3 Suppose \( f = u + iv \) is holomorphic in the region \( G \). Then \( u \) and \( v \) are harmonic in \( G \). **Proof:** First, by Corollary 5.5, \( u \) and \( v \) have continuous second partials. By Theorem 2.13, \( u \) and \( v \) satisfy the Cauchy–Riemann equations (2.3): \[ u_x = v_y \quad \text{and} \quad u_y = -v_x \] in \( G \). Hence we can repeat our argumentation in (2.4): \[ u_{xx} + u_{yy} = (u_x)_{x} + (u_y)_{y} = (v_y)_{x} + (-v_x)_{y} = v_{yx} - v_{xy} = 0. \] Note that in the last step we used the fact that \( v \) has continuous second partials. The proof that \( u \) satisfies the Laplace equation is practically identical. Proposition 6.3 gives us an effective way to show that certain functions are harmonic in \( G \) by way of constructing an accompanying holomorphic function on \( G \). ## Example 6.4 Revisiting Example 6.1, we can see that \( u(x,y) = x \) is harmonic in \( \mathbb{C} \) by noticing that \[ f(z) = \frac{1}{2} z^2 = \frac{1}{2} (x^2 - y^2) + i x y \] is entire and \( \text{Im}(f) = y \). ## Example 6.5 A second reason that the function \( u(x,y) \) from Example 6.2 is harmonic in \( C \) is that \[ f(z) = e^{r \cos \theta} + i e^{r \sin \theta} \] is entire and \( \text{Re}(f) = u \). Proposition 6.3 practically shows us a converse. There are, however, functions that are harmonic in a region \( G \) but not the real part (say) of a holomorphic function in \( G \) (Exercise 6.5). We do obtain a converse of Proposition 6.3 if we restrict ourselves to **simply-connected** regions. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 118 Context: # HARMONIC FUNCTIONS ## Theorem 6.6. Suppose \( u \) is harmonic on a simply-connected region \( G \). Then there exists a harmonic function \( v \) in \( G \) such that \( f = u + iv \) is holomorphic in \( G \). The function \( v \) is called a **harmonic conjugate of** \( u \). **Proof:** We will explicitly construct a holomorphic function \( f \) (and thus \( v = \text{Im } f \)). First, let \[ g = u_x - i u_y. \] The plan is to prove that \( g \) is holomorphic, and then to construct an antiderivative of \( g \), which will be almost the function \( f \) that we are after. To prove that \( g \) is holomorphic, we use Theorem 2.13: first because \( u \) is harmonic, \( \text{Re } g = u_x \) and \( \text{Im } g = -u_y \) have continuous partials. Moreover, again because \( u \) is harmonic, \( \text{Re } g \) and \( \text{Im } g \) satisfy the Cauchy-Riemann equations (2.3): \[ \begin{align*} \text{(Re } g)_y & = u_{xy} = -u_{yx} = -(\text{Im } g)_x, \\ \text{(Re } g)_x & = u_{xx} = -u_{yy} = -(\text{Im } g)_y. \end{align*} \] Theorem 2.13 implies that \( g \) is holomorphic in \( G \), and so we can use Corollary 5.8 to obtain an antiderivative \( f \) of \( g \) (here is where we use the fact that \( G \) is simply connected). Now we decompose \( g \) into its real and imaginary parts as \( h = a + ib \). Then, again using Theorem 2.13, \[ g = b' - a' i, \] (The second equation follows from the Cauchy-Riemann equations (2.3)). But the real part of \( g \) is \( g_x = a \), and thus \( u_x = a = \text{Re } f \) for some function \( f \) that depends only on \( y \). On the other hand, comparing the imaginary parts of \( g' \) yields \( -u_y = -a' = b_y = \text{Im } f \) where \( c \) depends only on \( y \). Hence \( f \) has the form \[ f(x) = h(x) + c \] is a function holomorphic in \( G \) whose real part is \( u \), as promised. > As a side remark, with hindsight it should not be surprising that the function \( g \) that we first constructed in our proof is the derivative of the sought-after function. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 119 Context: # Definition and Basic Properties Let \( f \). Namely, by Theorem 2.13 such a function \( f = u + iv \) must satisfy \[ f' = u_x + iv_x = u_y - iv_y. \] (The second equation follows from the Cauchy–Riemann equations (2.3).) It is also worth mentioning that our proof of Theorem 6.6 shows that if \( u \) is harmonic in \( G \), then \( v \) is the real part of the function \( g = u - iv \), which is holomorphic in \( G \) regardless of whether \( G \) is simply connected or not. As our proof of Theorem 6.6 is constructive, we can use it to produce harmonic conjugates. ## Example 6.7 Revisiting Example 6.1 for the second time, we can construct a harmonic conjugate of \( u(x, y) = xy \) along the lines of our proof of Theorem 6.6: first let \[ g = u_x - i v = y - i x = -i z, \] which has an antiderivative \[ h(z) = \frac{1}{2} z^2 = xy - \frac{1}{2}(x^2 - y^2) \] whose real part is \( u \) and whose imaginary part \[ v(x, y) = -\frac{1}{2}(x^2 - y^2) \] gives a harmonic conjugate for \( u \). We can give a more practical machinery for computing harmonic conjugates, which only depends on computing certain (calculus) integrals; thus this can be easily applied, e.g., to polynomials. We state it for functions that are harmonic in the whole complex plane, but you can easily adjust it to functions that are harmonic on certain subsets of \( \mathbb{C}^2 \). ## Theorem 6.8 Suppose \( u \) is harmonic on \( C \). Then \[ u(x, y) := \int \frac{\partial u}{\partial x}(x, t) \, dt - \int \frac{\partial u}{\partial y}(y, t) \, dt \] is a harmonic conjugate for \( u \). *Theorem 6.8 is due to Sheldon Axler and the basis for his Mathematica package Harmonic Function Theory.* #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 120 Context: ``` 114 # HARMONIC FUNCTIONS ## Proof: We will prove that \( u + iv \) satisfies the Cauchy–Riemann equations (2.3). The first follows from \[ \frac{\partial u}{\partial y}(x, y) = \frac{\partial v}{\partial x}(x, y) \] by the Fundamental Theorem of Calculus (Theorem A.3). Second, by Leibniz's Rule (Theorem A.9), the Fundamental Theorem of Calculus (Theorem A.3), and the fact that \( v \) is harmonic, \[ \frac{\partial v}{\partial x}(x, y) = \int_{0}^{y} \frac{\partial^2 v}{\partial x^2}(x, t) \, dt = -\int_{0}^{y} \frac{\partial^2 u}{\partial y^2}(x, t) \, dt = \frac{\partial u}{\partial y}(x, 0) - \frac{\partial^2 u}{\partial y \partial x}(x, 0). \] As you might imagine, Proposition 6.3 and Theorem 6.6 allow for a powerful interplay between harmonic and holomorphic functions. In that spirit, the following theorem appears not so surprising. You might appreciate its depth better when looking back at the simple definition of a harmonic function. ## Corollary 6.9 A harmonic function is infinitely differentiable. ## Proof: Suppose \( u \) is harmonic in \( G \) and \( z_0 \in G \). We wish to show that \( u \in C^\infty \) exists for all positive integers \( n \). Let \( r > 0 \) such that the disk \( D(z_0, r) \) is contained in \( G \). Since \( D(z_0, r) \) is simply connected, Theorem 6.6 asserts the existence of a holomorphic function \( f \) in \( D(z_0, r) \) such that \( \text{Re} \, f = u \) by Corollary 5.5. \( f \) is infinitely differentiable on \( D(z_0, r) \) and hence so is its real part \( u \). This proof is the first in a series of proofs that use the fact that the property of being harmonic is local—if it is a property at each point of a certain region. Note that in our proof of Corollary 6.9 we did not construct a function \( f \) that is holomorphic in \( G \); we only constructed such a function on the disk \( D(z_0, r) \). This might very well differ from one disk to the next. ## 6.2 Mean-Value and Maximum/Minimum Principle We have established an intimate connection between harmonic and holomorphic functions, and so it should come as no surprise that some of the theorems we proved for holomorphic functions have analogues in the world of harmonic functions. Here is such a harmonic analogue of Cauchy's Integral Formula (Theorems 6.24 and 6.27). ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 121 Context: Theorem 6.10. Suppose \( u \) is harmonic in the region \( G \) and \( \overline{D(u, r)} \subset G \). Then \[ u(x) = \frac{1}{2\pi} \int_{D(u,r)} u(y + r e^{i\theta}) d\theta. \] **Proof.** Exercise 6.14 provides \( x \in D(u,r) \subset D[u,r] \subset G \). The open disk \( D[u,r] \) is simply connected, so by Theorem 6.6 there is a function \( f \) holomorphic in \( D[u,r] \) such that \( u = \operatorname{Re} f \) on \( D[u,r] \). Now we apply Corollary 4.25 to \( f \): \[ f(x) = \frac{1}{2\pi} \int_{0}^{2\pi} f(x + r e^{i\theta}) d\theta. \] Theorem 6.10 follows by taking the real part on both sides. **Corollary** 4.25 and Theorem 6.10 say that holomorphic and harmonic functions have the mean-value property. Our next result is an important consequence of this property to extreme values of a function. **Definition.** Let \( G \) be a region. The function \( u : G \to \mathbb{R} \) has a strong relative maximum at \( x_0 \in G \) if there exists a disk \( D[u, r] \subset G \) such that \( u(x) \leq u(x_0) \) for all \( x \in D[u, r] \) and \( u(x_0) < u(x) \) for some \( y \in D[u, r] \). The definition of a strong relative minimum is analogous. Theorem 6.11. If \( u \) is harmonic in the region \( G \), then it does not have a strong relative maximum or minimum in \( G \). **Proof.** Assume, by way of contradiction, that \( u \) has a strong relative maximum. Then there is a disk in \( G \) centered at \( x_0 \) containing a point \( x \) with \( u(x) < u(x_0) \). Let \( r = |x - x_0| \) and apply Theorem 6.10: \[ u(x) = \frac{1}{2\pi} \int_{0}^{2\pi} u(x_0 + r e^{i\theta}) d\theta. \] Intuitively, this cannot hold, because some of the function values we're integrating are smaller than \( u(x_0) \), contradicting the mean-value property. To make this into a thorough argument, suppose that \( u(x) < u(x_0) \) for \( r \leq 5 \) and \( x \in [0, 2r] \). Now we split up the mean-value... #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 123 Context: # Mean-Value Principle functions, Corollary 8.20, concerns bounded regions. In Chapter 8 we will establish that, if \( u \) is harmonic in a bounded region \( G \) and continuous on its closure, then \[ \sup_{x \in G} u(x) = \max_{x \in \partial G} u(x) \quad \text{and} \quad \inf_{x \in G} u(x) = \min_{x \in \partial G} u(x) \tag{6.1} \] where, as usual, \( \partial G \) denotes the boundary of \( G \). We'll exploit this in the next two corollaries. ## Corollary 6.13 Suppose \( u \) is harmonic in the bounded region \( G \) and continuous on its closure. If \( u \) is zero on \( \partial G \) then it is zero in \( G \). **Proof.** By (6.1), \[ u(z) \leq \sup_{x \in G} u(x) = \max_{x \in \partial G} u(x) \] and \[ u(z) \geq \inf_{x \in G} u(x) = \min_{x \in \partial G} u(x), \] so \( u \) must be zero in \( G \. \) ## Corollary 6.14 Suppose \( u \) and \( v \) are harmonic in the bounded region \( G \) and continuous on its closure. If \( u(x) = v(x) \) for all \( x \in \partial G \) then \( u \equiv v \) for all \( x \in G \). **Proof.** \( u - v \) is harmonic in \( G \) (Exercise 6.2) and is continuous on the closure \( \overline{G} \), and \( u - v \) is zero on \( \partial G \). Now apply Corollary 6.13. ## Corollary 6.15 Corollary 6.14 says that if a function \( u \) is harmonic in a bounded region \( G \) and is continuous on the closure \( \overline{G} \) then the values of \( u \) at points in \( G \) are completely determined by the values of \( u \) on the boundary of \( G \). We should remark, however, that this result is of a completely theoretical nature: it says nothing about how to extend a continuous function \( f \) given on the boundary of a region to be harmonic in the full region. This problem is called the Dirichlet problem, and it has a solution for all bounded simply-connected regions. If the region \( G \) is the unit disk and \( f \) is a continuous function on the unit circle, define \[ \hat{u}(r, \theta) = r^{\ell} \cdot f(\theta) \quad \text{and} \quad \hat{u}(r, \theta) = \frac{1}{2\pi} \int_0^{2\pi} r^{\ell} f(\phi) \, d\phi \quad \text{for } r < 1, \] where \(\ell\) is a suitable integer. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 124 Context: where \( P_\nu(r) \) is the Poisson kernel which we introduced in Exercise 4.31. Then \( u \) is the desired extension: it is continuous on the closed unit disk, harmonic in the open unit disk, and agrees with \( u \) on the unit circle. In simple cases this solution can be converted to solutions in other regions, using a conformal map to the unit disk. All of this is beyond the scope of this book, though Exercise 6.13 gives some indication why the above formula does the trick. At any rate, we remark that Corollary 6.14 says that the solution to the Dirichlet problem is unique. ## Exercises 6.1. Show that all partial derivatives of a harmonic function are harmonic. 6.2. Suppose \( u(x, y) \) and \( v(x, y) \) are harmonic in \( G \), and \( \varepsilon \in \mathbb{R} \). Prove that \( u(x, y) + \varepsilon v(x, y) \) is also harmonic in \( G \). 6.3. Give an example that shows that the product of two harmonic functions is not necessarily harmonic. 6.4. Let \( u(x, y) = e^y \sin y \). (a) Show that \( u \) is harmonic on \( \mathbb{C} \). (b) Find an entire function \( f \) such that \( \text{Re}(f) = u \). 6.5. Consider \( u(x, y) = \ln(x^2 + y^2) \). (a) Show that \( u \) is harmonic on \( \mathbb{C} \setminus \{(0, 0)\} \). (b) Prove that \( u \) is not the real part of a function that is holomorphic in \( \mathbb{C} \setminus \{(0, 0)\} \). 6.6. Show that if \( f \) is holomorphic and nonzero in \( G \), then \( \ln|f(x, y)| \) is harmonic in \( G \). 6.7. Suppose \( u(x, y) \) is a function \( \mathbb{R}^2 \rightarrow \mathbb{R} \) that depends only on \( x \). When is \( u \) harmonic? #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 125 Context: 8. It is possible to find a real function \( r(x, y) \) so that \( x^2 + y^2 + r(x, y) \) is holomorphic? 6.9. Suppose \( f \) is holomorphic in the region \( G \subset \mathbb{C} \) with image \( H = \{ f(z) : z \in G \} \), and \( u \) is harmonic on \( G \). Show that \( u(f(z)) \) is harmonic on \( G \). 6.10. Suppose \( u(r, \phi) \) is a function \( \mathbb{R}^2 \to \mathbb{R} \) given in terms of polar coordinates. (a) Show that the Laplace equation for \( u(r, \phi) \) is \[ \frac{1}{r} \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial \phi^2} = 0. \] (b) Show that \( u(r, \phi) = r^2 \cos(2\phi) \) is harmonic on \( C \). Generalize. (c) If \( u(r, \phi) \) depends only on \( r \), when is it harmonic? (d) If \( u(r, \phi) \) depends only on \( \phi \), when is it harmonic? 6.11. Prove that if \( u \) is harmonic and bounded on \( C \), then it is constant. (Hint: Use Theorem 6.6 and Liouville's Theorem (Corollary 5.13).) 6.12. Suppose \( a(x, y) \) is a harmonic polynomial in \( x \) and \( y \). Prove that the harmonic conjugate of \( u \) is also a polynomial in \( x \) and \( y \). 6.13. Recall from Exercise 4.31 the Poisson kernel \[ P_\phi(r) = \frac{1 - r^2}{1 - 2r\cos(\phi) + r^2} \] where \( 0 < r < 1 \. In this exercise, we will prove the Poisson Integral Formula: If \( u \) is harmonic on an open set containing the closed unit disk \( D[0, 1] \) then for \( r < 1 \) \[ u(re^{i\phi}) = \frac{1}{2\pi} \int_0^{2\pi} P_\phi(r) u(1e^{i\theta}) \, d\theta. \quad (6.2) \] Suppose \( u \) is harmonic on an open set containing \( D[0, 1] \). By Exercise 6.14 we can find \( R_0 > 1 \) so that \( u \) is harmonic in \( D[R_0]. \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 126 Context: # Harmonic Functions (a) Recall the Möbius function \[ f(z) = \frac{z - a}{1 - \overline{a}z}, \] for some fixed \( a \in \mathbb{C} \) with \( |a| < 1 \), from Exercise 3.9. Show that \( u(f(z)) \) is harmonic on an open disk \( D[0, R] \) containing \( D(0, 1) \). (b) Apply Theorem 6.10 to the function \( u(f(z)) \) with \( v = 0 \) to deduce \[ u(a) = \frac{1}{2\pi} \int_{C(a)} \frac{u(f(z))}{z} dz. \] (c) Recalling, again from Exercise 3.9, that \( f(z) \) maps the unit circle to itself, apply a change of variables to (6.3) to prove \[ u(a) = \frac{1}{2\pi} \int_{0}^{2\pi} u(e^{i\theta}) \left| 1 - \frac{|a|^2}{|e^{i\theta} - a|^2} \right| d\theta. \] (d) Deduce (6.2) by setting \( a = re^{i\theta} \). ## Exercise 6.14 Suppose \( G \) is open and \( \overline{D(a, r)} \subset G \). Show that there is \( R > r \) so that \( \overline{D(a, r)} \subset D[R] \subset G \). (Hint: If \( G = \mathbb{C} \) just take \( R = r + 1 \). Otherwise choose some \( w \in G \), let \( M = |w - a| \), and let \( K = \overline{D(a, M)} \subset G \). Show that \( K \) is nonempty, closed, and bounded, and apply Theorem A1 to find a point \( x_0 \in K \) that minimizes \( f(z) = |z - a| \) on \( K \). Show that \( R = |x_0 - a| \) works.) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 129 Context: To prove that a sequence \((a_n)\) is divergent, we have to show the negation of the statement that defines convergence; that is, given any \(L \in \mathbb{C}\), there exists \(\epsilon > 0\) such that, given any integer \(N\), there exists an integer \(n > N\) such that \(|a_n - L| > \epsilon\). (If you have not negated any mathematical statements, this is worth meditating about.) ### Example 2.7: The sequence \((a_n = i^n)\) diverges: Given \(L \in \mathbb{C}\), choose \(L = \frac{1}{2}\). We consider two cases: If \(\text{Re}(L) > 0\), then for any \(N\), choose \(n > N\) such that \(a_n = -1\). (This is always possible since \(a_{2k} = i^{2k} = -1\) for any \(k \geq 0\).) Then \[ |a_n - L| = |1 + L| > \frac{1}{2}. \] If \(\text{Re}(L) < 0\), then for any \(N\), choose \(n > N\) such that \(a_n = 1\). (This is always possible since \(a_{4k} = i^{4k} = 1\) for any \(k \geq 0\).) Then \[ |a_n - L| = |1 - L| > \frac{1}{2}. \] This proves that \((a_n = i^n)\) diverges. The following limit laws are the cousins of the identities in Propositions 2.4 and 2.6, with one little twist. ### Proposition 7.3. Let \((a_k)\) and \((b_k)\) be convergent sequences and \(c \in \mathbb{C}\). Then 1. \(\lim_{k \to \infty} (a_k + b_k) = \lim_{k \to \infty} a_k + \lim_{k \to \infty} b_k.\) 2. \(\lim_{k \to \infty} (a_k - b_k) = \lim_{k \to \infty} a_k - \lim_{k \to \infty} b_k.\) 3. \(\lim_{k \to \infty} (c a_k) = c \cdot \lim_{k \to \infty} a_k.\) 4. \(\lim_{k \to \infty} (a_k b_k) = \lim_{k \to \infty} a_k \cdot \lim_{k \to \infty} b_k.\) 5. \(\lim_{k \to \infty} \frac{a_k}{b_k} = \frac{\lim_{k \to \infty} a_k}{\lim_{k \to \infty} b_k}\) where \(b_k \neq 0\) for all sufficiently large \(k\), assuming \(\lim_{k \to \infty} b_k \neq 0.\) 6. \(\lim_{k \to \infty} k = \infty.\) Again, the proof of this proposition is essentially a repeat from arguments we have given in Chapters 2 and 3, as you should convince yourself in Exercise 7.4. We will assume, as an axiom, that it is complete. To phrase this precisely, we need 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: 130 Context: # POWER SERIES ## Definition The sequence \((a_n)\) is monotonic if it is either nondecreasing \((a_n \geq a_{n-1}\) for all \(n)\) or nonincreasing \((a_n \leq a_{n-1}\) for all \(n)\). There are many equivalent ways of formulating the completeness property for the reals. Here is what we'll go by: ## Axiom (Monotone Sequence Property) Any bounded monotonic sequence converges. This axiom (or one of its many equivalent statements) gives arguably the most important property of the real number system: namely, that we can, in many cases, determine that a given sequence converges without knowing the value of the limit. In this sense we can use the sequence to define a real number. ## Example 7.4 Consider the sequence \((a_n)\) defined by \[ a_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}. \] This sequence is increasing (by definition) and each \(a_n \leq 3\) by Exercise 7.9. By the Monotone Sequence Property, \((a_n)\) converges, which allows us to define one of the most famous numbers in all of mathematics, \[ \epsilon = 1 + \lim_{n \to \infty} a_n. \] ## Example 7.5 Fix \(0 < r < 1\). We claim that \(\lim_{n \to \infty} r^n = 0\). First, the sequence \((a_n = r^n)\) converges because it is decreasing and bounded below by 0. Let \(L = \lim_{n \to \infty} r^n\). By Proposition 7.3, \[ L = \lim_{n \to \infty} r^n = r^1 = \lim_{n \to \infty} r^n = 0. \] Thus \((1 - r)L = 0\) and (since \(1 - r \neq 0\)) we conclude that \(L = 0\). We remark that the Monotone Sequence Property implies the Least Upper Bound Property: every nonempty set of real numbers with an upper bound has a least upper bound. The Least Upper Bound Property, in turn, implies the following theorem, which is often stated as a separate axiom: **Footnote:** Both the Archimedean Property and the Least Upper Bound Property can be used in (different) axiomatic developments of \(\mathbb{R}\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 132 Context: # Power Series A series converges to the limit (or sum) \( L \) by definition if \[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^{n} b_k = L. \] To prove that a series converges, we use the definition of the limit of a sequence: for any \( \epsilon > 0 \), we have to find an \( N \) such that for all \( n \geq N \), \[ \left| \sum_{k=1}^{n} b_k - L \right| < \epsilon. \] In the case of a convergent series, we usually write its limit as \( L = \sum_{k=1}^{\infty} b_k \) or \( L = \sum_{k=1}^{\infty} b_k \). ## Example 7.8 Fix \( z \in \mathbb{C} \) with \( |z| < 1 \). We claim that the geometric series \( \sum_{k=0}^{\infty} z^k \) converges with limit \[ \sum_{k=0}^{\infty} z^k = \frac{z}{1 - z}. \] In this case, we can compute the partial sums explicitly: \[ \sum_{k=0}^{n} z^k = z + z^2 + \cdots + z^{n} = \frac{z(1 - z^{n+1})}{1 - z}, \] whose limit as \( n \to \infty \) exists by Example 7.5, because \( |z| < 1 \). ## Example 7.9 Another series whose limit we can compute by manipulating the partial sums is \[ \sum_{k=1}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty} \sum_{k=1}^{n} \left( 1 - \frac{1}{k+1} \right). \] This simplifies to \[ = \lim_{n \to \infty} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{n} \right) = 1. \] A series where most of the terms cancel like this is called **telescoping**. Most of the time we can use the completeness property to check convergence of a series, and it is fortunate that the Monotone Sequence Property has a convenient translation into the language of series of real numbers. The partial sums of a series are often used to analyze convergence. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 135 Context: Proposition 7.17 (Integral Test). If \( f: [1, \infty) \to \mathbb{R} \) is continuous and nonincreasing, then \[ \int_{1}^{\infty} f(t) \, dt \leq \sum_{k=1}^{\infty} f(k) \leq \int_{1}^{\infty} f(t) \, dt + f(1). \] The Integral Test literally comes with a proof by picture—see Figure 7.2: the integral of \( f \) on the interval \([k, k + 1]\) is bounded between \( f(k) \) and \( f(k + 1) \). Adding the pieces gives the inequalities above for the partial sum versus the integrals from \( 1 \) to \( n \) and from \( 1 \) to \( n + 1 \), and the inequality persists in the limit. Corollary 7.18. If \( f: [1, \infty) \to \mathbb{R} \) is continuous and nonincreasing, then \( \sum_{k=1}^{\infty} f(k) \) converges if and only if \( \int_{1}^{\infty} f(t) \, dt \) is finite. Proof. Suppose \( \int_{1}^{\infty} f(t) \, dt = \infty \). Then the first inequality in Proposition 7.17 implies that the partial sums \( \sum_{k=1}^{n} f(k) \) are unbounded, and so Corollary 7.10 says that \( \sum_{k=1}^{\infty} f(k) \) cannot converge. Conversely, if \( \int_{1}^{\infty} f(t) \, dt \) is finite, then the second inequality in Proposition 7.17 says that the partial sums \( \sum_{k=1}^{n} f(k) \) are bounded; thus, again with Corollary 7.10, we conclude that \( \sum_{k=1}^{\infty} f(k) \) converges. Example 7.19. The series \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) converges for \( p > 1 \) and diverges for \( p \leq 1 \) (and the case \( p = 1 \) was the subject of Example 7.16) because \[ \int_{1}^{\infty} \frac{dx}{x^p} = \lim_{t \to \infty} \left( \frac{x^{1-p}}{1-p} \Big|_1^t + 1 \right) = \frac{1}{1 - p} \text{ if and only if } p > 1. \] By now you might be amused that we have collected several results on series whose terms are nonnegative real numbers. One reason is that such series are a bit easier to handle; another one is that there is a notion of convergence special to series that relates any series to one with only nonnegative terms. Definition. The series \( \sum_{k=1}^{\infty} a_k \) converges absolutely if \( \sum_{k=1}^{\infty} |a_k| \) converges. Theorem 7.20. If a series converges absolutely then it converges. This seems like an obvious statement, but its proof is, nevertheless, nontrivial. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 136 Context: ``` # 130 Power Series ## Proof Suppose \( \sum_{k=1}^{\infty} |\phi_k| \) converges. We first consider the case that each \( b_k \) is real. Let \[ b_k^* = \begin{cases} b_k & \text{if } b_k \geq 0, \\ 0 & \text{otherwise.} \end{cases} \] and \[ b_k^{**} = \begin{cases} 0 & \text{if } b_k < 0, \\ b_k & \text{otherwise.} \end{cases} \] Then \( 0 \leq b_k^* \leq |b_k| \) for all \( k \geq 1 \), and so by Corollary 7.12, both \[ \sum_{k=1}^{\infty} b_k^* \] and \[ -\sum_{k=1}^{\infty} b_k^{**} \] converge. But then so does \[ \sum_{k=1}^{\infty} b_k = \sum_{k=1}^{\infty} b_k^* + \sum_{k=1}^{\infty} b_k^{**}. \] For the general case \( b_k \in \mathbb{C} \), we write each term as \( b_k = c_k + i d_k \). Since \( 0 \leq |c_k| \leq |b_k| \) for all \( k \geq 1 \), Corollary 7.12 implies that \( \sum_{k=1}^{\infty} |c_k| \) converges absolutely, and by an analogous argument, so does \( \sum_{k=1}^{\infty} |d_k| \). But now we can use the first case to deduce that both \( \sum_{k=1}^{\infty} c_k \) and \( \sum_{k=1}^{\infty} d_k \) converge. ### Example 7.21. Continuing Example 7.19, we find: \[ \zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k^s} \] converges for \( \text{Re}(s) > 1 \), because then (using Exercise 3.49) \[ \sum_{k=1}^{\infty} k^{-s} = \sum_{n=1}^{\infty} n^{-\text{Re}(s)} \] converges. Viewed as a function in \( s \), the series \( \zeta(s) \) is the Riemann zeta function, an indispensable tool in number theory and many other areas in mathematics and physics. > **Note:** The Riemann zeta function is the subject of the subtly most famous open problem in mathematics, the **Riemann Hypothesis**. It turns out that \( \zeta(s) \) can be extended to a function that is holomorphic on \( \mathbb{C} \setminus \{1\} \), and the Riemann hypothesis asserts that the roots of this extended function in the strip \( 0 < \text{Re}(s) < 1 \) are all on the critical line \( \text{Re}(s) = \frac{1}{2} \). ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 137 Context: Another common mistake is to try to use the converse of Theorem 7.20, which is also false: ### Example 7.22 The alternating harmonic series \[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} \] converges: \[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \] \[ = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \ldots \] (There is a small technical detail to be checked here, since we are effectively ignoring half the partial sums of the original series; see Exercise 7.16.) Since \[ \frac{1}{2k - 1} - \frac{1}{2k} \leq \frac{1}{2k - 1} \leq \frac{1}{k^2} \] \[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} \text{ converges by Corollary 7.12 and Example 7.19.} \] However, according to Example 7.16, \[ \sum_{k=1}^{\infty} \frac{1}{k} \text{ does not converge absolutely.} \] ## 7.3 Sequences and Series of Functions The fun starts when we study sequences of functions. ### Definition Let \( G \subset \mathbb{C} \) and \( f_n : G \to C \) for \( n \geq 1 \). We say that \( (f_n) \) converges pointwise to \( f : G \to \mathbb{C} \) if for each \( x \in G \), \[ \lim_{n \to \infty} f_n(x) = f(x). \] We say that \( (f_n) \) converges uniformly to \( f : G \to \mathbb{C} \) if there is an \( N \) such that for all \( x \in G \) and all \( n \geq N \), \[ |f_n(z) - f(z)| < \epsilon. \] Sometimes we want to express that either notion of convergence holds only on a subset \( H \) of \( G \), in which case we say that \( (f_n) \) converges pointwise/uniformly on \( H \). It should be clear that uniform convergence on a set implies pointwise convergence on that set; but the converse is not generally true. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 139 Context: # Sequences and Series of Functions ## Proposition 7.25 Suppose \( C \) and \( f_n : G \to C \) is continuous, for each \( n \geq 1 \). If \( f_n \) converges uniformly to \( f : G \to C \) then \( f \) is continuous. **Proof:** Let \( x_0 \in G \); we will prove that \( f \) is continuous at \( x_0 \). By uniform convergence, given \( \epsilon > 0 \), there is an \( N \) such that for all \( x \in G \) and all \( n \geq N \): \[ |f_n(x) - f(x_0)| < \frac{\epsilon}{5} \] Now we make use of the continuity of the \( f_n \)'s. This means that given (the same) \( \epsilon > 0 \), there is a \( \delta > 0 \) such that whenever \( |x - x_0| < \delta \): \[ |f_n(x) - f_n(x_0)| < \frac{\epsilon}{5} \] All that’s left is to put those two inequalities together: by the triangle inequality (Corollary 1.7(1)), \[ |f(x) - f(x_0)| = |f(x) - f_n(x) + f_n(x) - f_n(x_0) + f_n(x_0) - f(x_0)| \] \[ \leq |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \frac{\epsilon}{5} + \frac{\epsilon}{5} + \frac{\epsilon}{5} = \frac{3\epsilon}{5} < \epsilon \] This proves that \( f \) is continuous at \( x_0 \). ## Proposition 7.25 Proposition 7.25 can sometimes give a hint that a function sequence does not converge uniformly. ## Example 7.26 We modify Example 7.23 and consider the real function sequence \( f_n : [0, 1] \to \mathbb{R} \) given by \[ f_n(x) = \begin{cases} 0 & \text{if } 0 \leq x < 1, \\ 1 & \text{if } x = 1. \end{cases} \] As this limiting function is not continuous, the above convergence cannot be uniform. This gives a strong indication that the convergence in Example 7.23 is not uniform. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 140 Context: # POWER SERIES uniform either, though this needs a separate proof, as the domain of the functions in Example 7.23 is the unit disk (Exercise 7.20(b)). Now that we have established Proposition 7.25 about continuity, we can ask about integration of sequences or series of functions. The next theorem should come as no surprise; however, its consequences (which we will see shortly) are wide ranging. ## Proposition 7.27 Suppose \( f_n : G \to \mathbb{C} \) is continuous, for \( n \geq 1 \). Then \( (f_n) \) converges uniformly to \( f : G \to \mathbb{C} \) and \( G \) is a piecewise smooth path. Then \[ \lim_{n \to \infty} \int_{C} f_n(z) \, dz = \int_{C} f(z) \, dz. \] **Proof:** We may assume that \( z \) is not just a point, in which case the proposition holds trivially. Given \( \epsilon > 0 \), there exists \( N \) such that for all \( z \in G \) and all \( n \in \mathbb{N}, \) \[ |f_n(z) - f(z)| < \frac{\epsilon}{\text{length}(C)}. \] With Proposition 4.6(d) we can thus estimate \[ \left| \int_{C} f_n(z) \, dz - \int_{C} f(z) \, dz \right| \leq \max_{z \in C} |f_n(z) - f(z)| \cdot \text{length}(C) < \epsilon. \] All of these notions for sequences of functions hold verbatim for series of functions. For example, if \( \sum_{n} f_n(z) \) converges uniformly on \( G \) and \( G \) is a piecewise smooth path, then \[ \int_{C} \sum_{n} f_n(z) \, dz = \sum_{n} \int_{C} f_n(z) \, dz. \] In some sense, the above identity is why we care about uniform convergence. There are several criteria for uniform convergence; see, e.g., Exercises 7.19 and 7.20, and the following result, sometimes called the Weierstrass M-test. ## Proposition 7.28 Suppose \( f_n : G \to \mathbb{C} \) for \( n \geq 1 \), and \( |f_n(z)| \leq M_n \) for all \( z \in G \), where \( \sum_{n} M_n \) converges. Then \( \sum_{n} f_n(z) \) converges absolutely and uniformly in \( G \). (We say the series \( \sum_{n} f_n(z) \) converges absolutely and uniformly.) **Proof:** For each fixed \( z \), the series \( \sum_{n} |f_n(z)| \) converges absolutely by Corollary 7.12. To show that the convergence is uniform, let \( \epsilon > 0 \). Then there exists \( N \) such that #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 142 Context: # Example 7.30 A slight modification of Example 7.29 gives a fundamental power series, namely the geometric series \[ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}. \] So here \(a_0 = 0\) and \(c_k = 1\) for all \(k \geq 0\). We note that, as in Example 7.29, this power series converges absolutely for \(|x| < 1\) and uniformly for \(|x| \leq r\), for any fixed \(r < 1\). Finally, as in Example 7.15, the geometric series \(\sum_{k=0}^{\infty} x^k\) diverges for \(|x| \geq 1\). A general power series has a very similar convergence behavior which, in fact, comes from comparing it to a geometric series. ## Theorem 7.31 Given a power series \(\sum_{n=0}^{\infty} c_n (x - a)^n\), there exists a real number \(R \geq 0\) or \(R = \infty\), such that 1. \(\sum_{n=0}^{\infty} c_n (x - a)^n\) converges absolutely for \(|x - a| < R\); 2. \(\sum_{n=0}^{\infty} c_n (x - a)^n\) converges absolutely and uniformly for \(|x - a| \leq r\), for any \(r < R\); 3. \(\sum_{n=0}^{\infty} c_n (x - a)^n\) diverges for \(|x - a| > R\). We remark that this theorem says nothing about the convergence/divergence of \(\sum_{n=0}^{\infty} c_n (x - a)^n\) for \(|x - a| = R\). ## Definition The number \(R\) in Theorem 7.31 is called the **radius of convergence** of \(\sum_{n=0}^{\infty} c_n (x - a)^n\). The open disk \(D(a,R)\) in which the power series converges absolutely is the region of convergence. (If \(R = 0\) then this is \(C\).) In preparation for the proof of Theorem 7.31, we start with the following observation. ## Proposition 7.32 \(\sum_{n=0}^{\infty} c_n (x - a)^n\) converges then \(\sum_{n=0}^{\infty} c_n (y - a)^n\) converges absolutely whenever \(|x - a| < |y - a|\). ### Proof Let \(r = |x - a|\). If \(\sum_{n=0}^{\infty} c_n (w - a)^n\) converges then \(\lim_{n \to \infty} c_n (w - a)^n = 0\) and so this sequence of terms is bounded (by Exercise 7.6), say \[ |c_n (w - a)^n| = |x - a|^n \leq M. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 146 Context: # Exercises ## 7.1 For each of the sequences, prove convergence or divergence. If the sequence converges, find the limit. (a) \( a_n = e^{n} \) (b) \( a_n = \frac{1}{n} \) (c) \( a_n = \cos(n) \) (d) \( a_n = \sin\left(\frac{1}{n}\right) \) (e) \( a_n = 2 - \frac{n^2}{2n^2+1} \) ## 7.2 Determine whether each of the following series converges or diverges. (a) \( \sum_{n=2}^{\infty} \left(\frac{1}{n}\right)^{\frac{3}{2}} \) (b) \( \sum_{n=2}^{\infty} \left(\frac{1}{n}\right)^{1} \) (c) \( \sum_{n=2}^{\infty} \left( \frac{1 + 2i}{\sqrt{5}} \right) \) (d) \( \sum_{n=2}^{\infty} \frac{1}{n^2 + 2n} \) ## 7.3 Compute \( \sum_{n=2}^{\infty} \frac{1}{n + 2n} \). ## 7.4 Prove Proposition 7.3. ## 7.5 Prove the following: (a) \( \lim_{n \to \infty} a_n = L \iff \lim_{n \to \infty} |a_n| = |L| \). (b) \( \lim_{n \to \infty} a_n = 0 \implies \lim_{n \to \infty} |a_n| = 0 \). ## 7.6 Show that a convergent sequence is bounded, i.e.: if \( \lim_{n \to \infty} a_n = L \) exists, then there is an \( M \) such that \( |a_n| \leq M \) for all \( n \geq 1 \). ## 7.7 Show that the limit of a convergent sequence is unique. ## 7.8 Let \( (a_n) \) be a sequence. A point \( a \) is an accumulation point of the sequence if for every \( \epsilon > 0 \) and every \( N \in \mathbb{Z}^+ \) there exists some \( n > N \) such that \( |a_n - a| < \epsilon \). Prove that if a sequence has more than one accumulation point then the sequence diverges. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 147 Context: # REGIONS OF CONVERGENCE ## 7.9 (a) Show that \( \frac{1}{3} \sum_{k=1}^{\infty} \frac{1}{k} \) for any positive integer \( k \). (b) Conclude with Example 7.9 that for any positive integer \( n \), \[ 1 + \frac{1}{2} + \frac{1}{6} + \ldots + \frac{1}{n!} \geq 3. \] ## 7.10 Derive the Archimedean Property from the Monotone Sequence Property. ## 7.11 Prove Proposition 7.7. ## 7.12 Prove that \( (c_n) \) converges if and only if \( (\text{Re} c_n) \) and \( (\text{Im} c_n) \) converge. ## 7.13 Prove that \( Z \) is complete and that \( Q \) is not complete. ## 7.14 Prove that, if \( a_n \leq b_n \leq c_n \) for all \( n \) and \( \lim_{n \to \infty} a_n = L, \, b_n = \lim_{n \to \infty} b_n = L, \, c_n = L \), then \( \lim_{n \to \infty} b_n = L \). This is called the Squeeze Theorem, and is useful in testing a sequence for convergence. ## 7.15 Find the least upper bound of the set \( \{ \text{Re}(e^{(2\pi i t)}) : t \in Q \} \). ## 7.16 (a) Suppose that the sequence \( (c_n) \) converges to zero. Show that \( \sum_{n=0}^{\infty} c_n \) converges if and only if \( \sum_{n=0}^{\infty}(c_n + c_{n+1}) \) converges. Moreover, if the two series converge then they have the same limit. (b) Give an example where \( (c_n) \) does not converge to 0 and one of the series in (a) diverges while the other converges. ## 7.17 Prove that the series \( \sum b_n \) converges if and only if \( \lim_{n \to \infty} \sum_{k=0}^{n} b_k = 0 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 149 Context: # REGIONS OF CONVERGENCE ## 7.21 Consider \( f_n : [0, \pi] \to \mathbb{R} \) given by \( f_n(x) = \sin(x/n) \) for \( n \geq 1 \). Prove that \( (f_n) \) converges pointwise to \( f : [0, \pi] \to \mathbb{R} \) given by \[ f(x) = \begin{cases} 1 & \text{if } x = 0 \\ 0 & \text{if } x \neq 0 \end{cases} \] yet this convergence is not uniform. (See Figure 7.3.) ## 7.22 Where do the following sequences converge pointwise? Do they converge uniformly on this domain? - (a) \( (a_n) \) - (b) \( \left(\frac{1}{n}\right) \) - (c) \( \left(\frac{1}{z}\right) \) where \( \text{Re}(z) \geq 0 \) ## 7.23 Let \( f(x) = x^k e^{-x} \). (a) Show that \( \lim_{n \to \infty} n f_n(x) = 0 \) for all \( x \geq 0 \). (Hint: Treat \( x = 0 \) as a special case; for \( x > 0 \), you can use L'Hôpital's rule (Theorem A.11) — but remember that \( x \) is the variable, not \( n \).) (b) Find \( \lim_{n \to \infty} \int_0^\infty f_n(x) \, dx \). (Hint: The answer is not \( 0 \).) (c) Why doesn't your answer to part (b) violate Proposition 7.27? ## 7.24 The product of two power series centered at \( x_0 \) is another power series centered at \( x_0 \). Derive a formula for its coefficients in terms of the coefficients of the original two power series. ## 7.25 Find a power series (and determine its radius of convergence) for the following functions. 1. \( \frac{1}{1 + 4x} \) 2. \( \frac{1}{3 - x} \) 3. \( \frac{x^2}{(4 - 2x)} \) ## 7.26 Find a power series representation about the origin of each of the following functions. - (a) \( \cos x \) - (b) \( \cos(z^2) \) - (c) \( z^2 \sin z \) - (d) \( \sin^2 x \) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 153 Context: # POWER SERIES AND HOLOMORPHIC FUNCTIONS A special case of this result concerns power series with infinite radius of convergence: those represent entire functions. Now that we know that power series are differentiable in their regions of convergence, we can also know how to find their derivatives. The next result says that we can similarly differentiate the series term by term. ## Theorem 8.2 Suppose \( f(z) = \sum_{k=0}^{\infty} c_k (z - z_0)^k \) has radius of convergence \( R > 0 \). Then \[ f'(z) = \sum_{k=1}^{\infty} k c_k (z - z_0)^{k-1} \quad \text{for any } z \in D_{[z_0,R]}. \] The radius of convergence of this power series is also \( R \). **Proof.** If \( z \in D_{[z_0,R]} \) then \( |z - z_0| < R \), so we can choose \( R' \) such that \( |z - z_0| < R' < R \). Then the circle \( \gamma = C(z_0,R') \) lies in \( D_{[z_0,R]} \) and is inside \( S \). Since \( f \) is holomorphic in \( D_{[z_0,R]} \), we can use Cauchy’s Integral Formula for \( f \) (Theorem 5.1), as well as Corollary 3.7: \[ f'(z) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(w)}{(w - z)^2} \, dw = \frac{1}{2 \pi i} \int_{\gamma} \frac{(w - z)^k}{(w - z)^2} \, dw = \sum_{k=0}^{\infty} k c_k \int_{\gamma} (w - z)^{k-1} \, dw \] Note that we used Theorem 5.1 again in the penultimate step, but now applied to the function \( f(w) \). The last statement of the theorem is easy to show: the radius of convergence of \( f'(z) \) is at least \( R \) since we have shown that the series for \( f' \) converges whenever \( |z - z_0| < R \), and it cannot be larger than \( R \) by comparison to the series for \( f(z) \), since the coefficients for \( (z - z_0)^{k-1} \) are larger than the corresponding ones for \( f(z) \). ## Example 8.3 Let \[ f(z) = \sum_{k=0}^{\infty} \frac{z^k}{k!}. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 155 Context: # POWER SERIES AND HOLOMORPHIC FUNCTIONS ## Proof For starters, \( f(0) = c_0 \). Theorem 8.2 gives \( f'(0) = c_1 \). Applying the same theorem to \( f' \) gives \[ f''(z) = \sum_{k=2}^{\infty} k(k-1)c_kz^{k-2} \] and so \( f''(c_0) = 2c_2 \). A quick induction game establishes \( f^{(n)}(0) = n!c_n \) for all \( k \geq 0 \). Taylor's formula shows that the coefficients of any power series converging to \( f \) on some open disk \( D \) can be determined from the function \( f \) restricted to \( D \). It follows immediately that the coefficients of a power series are unique: ### Corollary 8.6 If \( \sum_{k=0}^{\infty} a_k(z - z_0)^k \) and \( \sum_{k=0}^{\infty} b_k(z - z_0)^k \) are two power series that both converge to the same function on an open disk centered at \( z_0 \), then \( a_k = b_k \) for all \( k \geq 0 \). ### Example 8.7 We'd like to compute a power series expansion for \( f(z) = \exp(z) \) centered at \( z_0 = 0 \): \[ f^{(k)}(0) = \frac{1}{k!} \exp(0) = 1. \] Corollary 8.5 suggests that this power series is \[ \sum_{k=0}^{\infty} \frac{z^k}{k!} \] which converges for all \( z \in \mathbb{C} \) (essentially by Example 7.35). We now turn to the second cornerstone result of this section, that a holomorphic function can be locally represented by a power series. ### Theorem 8.8 Suppose \( f \) is a function holomorphic in \( D\left(z_0, r\right) \). Then \( f \) can be represented as a power series centered at \( z_0 \) with a radius of convergence \( r \): \[ f(z) = \sum_{k=0}^{\infty} c_k(z - z_0)^k \] with \[ c_k = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}} \, dz \] where \( \gamma \) is any positively oriented, simple, closed, piecewise smooth path in \( D\left(z_0, r\right) \) for which \( z_0 \) is inside \( \gamma \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 157 Context: # POWER SERIES AND HOLOMORPHIC FUNCTIONS What we have proved above, on the side, is the following. ## Corollary 8.9 If \( f: G \to \mathbb{C} \) is holomorphic and \( z_0 \in G \), then \( f \) can be expanded into a power series centered at \( z_0 \) whose radius of convergence is at least the distance of \( z_0 \) to \( \partial G \). ## Example 8.10 Consider \( f: \mathbb{C} \setminus \{1\} \to \mathbb{C} \) given by \( f(z) = \frac{1}{z+1} \) and \( z_0 = 0 \). Corollary 8.9 says that the power series expansion of \( f \) at \( 0 \) will have radius of convergence \( 1 \) (Actually, it says this radius is at least \( 1 \), but it cannot be larger since \( z = 1 \) are singularities of \( f \)). In fact, we can use a geometric series to compute this power series: \[ f(z) = \frac{1}{2} \cdot \frac{1}{1 - (-z)} = \sum_{n=0}^\infty (-z)^n = \sum_{n=0}^\infty (-1)^n z^n \] with radius of convergence \( 1 \). ## Corollary 8.11 Here we present yet another example of a result that is plainly false when translated into \( \mathbb{R} \); see Exercise 8.6. Comparing the coefficients of the power series obtained in Theorem 8.8 with those in Corollary 8.5, we arrive at the long-promised extension of Theorems 4.27 and 5.1. Suppose \( f \) is holomorphic in the region \( G \) and \( w \) is a positively oriented, simple, closed, piecewise smooth path, such that \( w \) is inside \( r \) and \( r \to 0 \). Then \[ \hat{f}(w) = \frac{1}{2\pi i} \int_{w} \frac{f(z)}{(z - w)^n} \, dz. \] Corollary 8.11 combined with our often-used Proposition 4.6(b) gives an inequality which is often called Cauchy’s Estimate. ## Corollary 8.12 Suppose \( f \) is holomorphic in \( D[R] \) and \( |f(z)| \leq M \) for all \( z \in D[R] \). Then \[ | \hat{f}(w) | \leq \frac{K M}{R^k}. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 159 Context: # Classification of Zeros and the Identity Principle Continuing in this way, we see that we can factor \( p(z) \) as \( p(z) = (z-a)^m g(z) \) where \( m \) is a positive integer and \( g(z) \) is a polynomial that does not have a zero at \( a \). The integer \( m \) is called the multiplicity of the zero of \( p(z) \). Almost exactly the same thing happens for holomorphic functions. ## Theorem 8.14 (Classification of Zeros) Suppose \( f : G \to \mathbb{C} \) is holomorphic and \( f \) has a zero at \( a \in G \). Then either: 1. \( f \) is identically zero on some disk \( D \) centered at \( a \) (that is, \( f(z) = 0 \) for all \( z \in D \)); or 2. There exists a positive integer \( m \) and a holomorphic function \( g : G \to \mathbb{C} \), such that \( f(z) = (z-a)^m g(z) \) for all \( z \in G \). In this case, the zero is isolated: there is a disk \( D[a,r] \) which contains no other zero of \( f \). The integer \( m \) in the second case is uniquely determined by \( f \) and is called the multiplicity of the zero at \( a \). ### Proof By Theorem 8.8, there exists \( R > 0 \) such that we can expand \[ f(z) = \sum_{k=0}^{\infty} c_k (z-a)^k \text{ for } z \in D[a,R]. \] and \( c_0 = f(a) = 0 \). There are now exactly two possibilities: either 1. \( c_0 = 0 \) for all \( k \geq 0 \); or 2. There is some positive integer \( m \) so that \( c_k = 0 \) for all \( k < m \) but \( c_m \neq 0 \). The first case gives \( f(z) = 0 \) for all \( z \in D[a,R] \). So now consider the second case. We note that for \( z \in D[a,R] \): \[ f(z) = (z-a)^m \left( c_m + c_{m+1} (z-a) + \ldots \right) = (z-a)^m \sum_{n=0}^{\infty} c_{m+n} (z-a)^n. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 160 Context: # TAYLOR AND LAURENT SERIES Thus we can define a function \( g : G \rightarrow \mathbb{C} \) through \( g(z) = \begin{cases} \sum_{n=0}^{\infty} a_n (z - a)^n & \text{if } z \in D(a, R), \\ f(z) & \text{if } z \in G \setminus \{ a \}. \end{cases} \) (According to our calculations above, the two definitions give the same value when \( z \in D(a, R) \). The function \( g \) is holomorphic in \( D(a, R) \) by the first definition, and \( g \) is holomorphic in \( G \setminus \{ a \} \) by the second definition. Note that \( g(a) = c \neq 0 \) and, by construction, \( f(z) = (z - a)^n s(z) \quad \text{for all } z \in G. \) Since \( s(a) \neq 0 \) by continuity, \( r > 0 \) so that \( s(z) \neq 0 \) for all \( z \in D(a, r) \), so \( D(a, r) \) contains no other zero of \( f \). The integer \( r \) is unique, since it is defined in terms of the power series expansion of \( f \) at \( a \), which is unique by Corollary 8.6. Theorem 8.14 gives rise to the following result, which is sometimes called the identity principle or the uniqueness theorem. ## Theorem 8.15 Suppose \( G \) is a region, \( f : G \rightarrow \mathbb{C} \) is holomorphic, and \( f(a) = 0 \) where \( (a_k) \) is a sequence of distinct numbers that converges in \( G \). Then \( f \) is the zero function on \( G \). Applying this theorem to the difference of two functions immediately gives the following variant. ## Corollary 8.16 Suppose \( f \) and \( g \) are holomorphic in a region \( G \) and \( f(a) = g(a) = 0 \) at a sequence that converges to \( a \in G \) with \( a_k \neq a \) for all \( k \). Then \( f(z) = g(z) \) for all \( z \in G \). ### Proof of Theorem 8.15 Consider the following two subsets of \( G \): - \( X := \{ z \in G : \text{ there exists } z \text{ such that } F(z) = 0 \text{ for all } z \in D(a, r) \} \) - \( Y := \{ z \in G : \text{ there exists } z \text{ such that } F(z) \neq 0 \text{ for all } z \in D(a, r) \setminus \{ a \} \} \) If \( f(a) \neq 0 \) then, by continuity of \( f \), there exists a disk centered at \( a \) in which \( f \) is nonzero, and so \( a \in Y \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 161 Context: # CLASSIFICATION OF ZEROS AND THE IDENTITY PRINCIPLE If \( f(z) = 0 \), then Theorem 8.14 says that either \( z \in X \) or \( z \) is an isolated zero of \( f \), so \( z \in V \). We have thus proved that \( G \) is the disjoint union of \( X \) and \( Y \). Exercise 8.11 proves that \( X \) and \( Y \) are open, and so (because \( G \) is a region) either \( X \) has to be empty. The conditions of Theorem 8.15 say that \( z_1, z_2, \ldots \) is not in \( V \), and this has to be in \( X \). Thus \( G = X \) and the statement of Theorem 8.15 follows. The identity principle yields the strengthenings of Theorem 6.11 and Corollary 6.12 promised in Chapter 6. We recall that we say the function \( f: G \to \mathbb{C} \) has a weak relative maximum if there exists a disk \( D(z_r, r) \subset G \) such that all \( z \in D(z_r, r) \) satisfy \( |f(z)| \leq |f(z_r)| \). ## Theorem 8.17 (Maximum-Modulus Theorem). Suppose \( f \) is holomorphic and nonconstant in a region \( G \). Then \( |f| \) does not attain a weak relative maximum in \( G \). There are many reformulations of this theorem, such as: ### Corollary 8.18. Suppose \( f \) is holomorphic in a bounded region \( G \) and continuous on its closure. Then \[ \sup_{z \in \partial G} |f(z)| = \max_{z \in \overline{G}} |f(z)|. \] **Proof.** This is trivial if \( f \) is constant, so we assume \( f \) is non-constant. By the Extreme Value Theorem, if there is a point \( z_r \in G \) such that \( \max_{z \in \overline{G}} |f(z)| \). Clearly \( \sup_{z \in |f(z)|} \geq \max_{z \in |f(z)|} \); and this is easily seen to be an equality using continuity at \( z_r \) since there are points of \( G \) arbitrarily close to \( z_r \). Finally, Theorem 8.17 implies \( f \) is not constant in \( G \). ### Corollary 8.19 (Minimum-Modulus Theorem). Suppose \( f \) is holomorphic and nonconstant in a region \( G \). Then \( |f| \) does not attain a weak relative minimum at a point \( z \in G \) unless \( f(z) = 0 \). ### Corollary 8.20. If \( f \) is harmonic and nonconstant in a region \( G \), then it does not have a weak relative maximum or minimum in \( G \). Note that Equation (6.1) in Chapter 6 follows from Corollary 8.20 using essentially the same argument as in the proof of Corollary 8.18. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 162 Context: # TAYLOR AND LAURENT SERIES ## Proof of Theorem 8.17 Suppose there exist \( G \) and \( R > 0 \) such that \( |f(z)| \geq |f(0)| \) for all \( z \in D(0, R) \). We will show that \( f \) is constant. If \( f(0) = 0 \) then \( f(z) = 0 \) for all \( z \in D(0, R) \) by Theorem 8.15. Now assume \( f(0) \neq 0 \), which allows us to define the holomorphic function \( g: G \to \mathbb{C} \) as \( g(z) = \frac{f(z)}{f(0)} \). This function satisfies \[ |g(z)| \leq |z| \quad \text{for all } z \in D(0, R). \] Also, \( g(0) = 1 \), so by continuity of \( g \), we can find \( r < R \) such that \( \text{Re}(g(z)) > 0 \) for \( z \in D(0, r) \). This allows us, in turn, to define the holomorphic function \( h: D(0, r) \to \mathbb{C} \) through \( h(z) = \log(g(z)) \), which satisfies \[ h(0) = \log(g(0)) = \log(1) = 0 \] and \[ \text{Re}(h(z)) = \text{Re}(\log(g(z))) \leq \ln|g(z)| \leq 0. \] Exercise 8.35 now implies that \( h \) must be identically zero in \( D(0, r) \). Hence \( g(z) = \exp(h(z)) \) must be equal to \( \exp(0) = 1 \) for all \( z \in D(0, r) \), and so \( f(z) = f(0) \) must have the constant value \( f(a) \) for all \( a \in D(0, r) \). Corollary 8.16 then implies that \( f \) is constant in \( G \). ## 8.3 Laurent Series Theorem 8.8 gives a powerful way of describing holomorphic functions. It is, however, not as general as it could be. It is natural, for example, to think about representing \( \exp(z) \) as \[ \exp\left(\frac{1}{z}\right) = \sum_{k=0}^{\infty} \left( \frac{1}{k!} \right) \cdot z^{-k}. \] a "power series" with negative exponents. To make sense of expressions like the above, we introduce the concept of a double series: \[ \sum_{k \in \mathbb{Z}} a_k = \sum_{k \geq 0} a_k + \sum_{k < 0} a_k. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 165 Context: # LAURENT SERIES ## Theorem 8.24 Suppose \( f \) is a function that is holomorphic in \( A = \{ z \in \mathbb{C} : R_1 < |z - z_0| < R_2 \} \). Then \( f \) can be represented in \( A \) as a Laurent series centered at \( z_0 \): \[ f(z) = \sum_{n=-\infty}^{\infty} c_n (z - z_0)^n \] with \[ c_n = \frac{1}{2 \pi i} \int_{C_{r_1}} \frac{f(w)}{(w - z_0)^{n+1}} \, dw, \] where \( R_1 < r < R_2 \). By Cauchy's Theorem 4.18, we can replace the circle \( C_{r_1} \) in the formula for the Laurent coefficients by any path \( \gamma \subset A [C_{r_1}, r_2] \). ![Figure 8.1: The path γ in our proof of Theorem 8.24.](path_to_figure) ## Proof Let \( g(z) = f(z + z_0) \), so \( g \) is a function holomorphic in \( \{ z \in \mathbb{C} : R_1 < |z| < R_2 \} \). Fix \( R_1 < r_1 < r_2 < R_2 \), and let \( \gamma \) be the path in Figure 8.1, where \( \gamma_1 = C[r_1, r_2] \) and \( \gamma_2 = C[0, r_2] \). By Cauchy's Integral Formula (Theorem 4.27), \[ g(z) = \frac{1}{2 \pi i} \int_{\gamma} \frac{g(w)}{w - z} \, dw. \tag{8.1} \] Complete any missing details from the original image as necessary while ensuring the accuracy of the markup. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 168 Context: ```markdown # TAYLOR AND LAURENT SERIES By Examples 8.3 and 8.23, \[ \frac{\exp(z)}{\sin(z)} = \left( 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \right) \left( -\frac{1}{z} + \frac{7}{360}z + \frac{31}{15120}z^3 + \cdots \right) = z^{-1} + 1 + \frac{2}{3} z + \cdots \] and Corollary 8.27 gives \[ \int_{C[0,1]} \frac{\exp(z)}{\sin(z)} \, dz = 2\pi i. \] For the integral around \( C \), we use the fact that \( \sin(-z) = -\sin(z) \), and so we can compute the Laurent expansion of \( \frac{\exp(z)}{\sin(z)} \) at \( z = 0 \) as Example 8.23: \[ \frac{1}{\sin(z)} = -\frac{1}{z} - \frac{1}{6} (z - z^3) - \frac{7}{360} (z - z^3) - \cdots. \] Adding Example 8.7 to the mix yields \[ \frac{\exp(z)}{\sin(z)} = \left( z^{-1} + \frac{z^{-1}}{2} + \frac{z^{-1}}{2} + \cdots \right) - \left( -\frac{1}{6}(z - z^3) - \cdots \right) \] \[ = -z^{-1} - \frac{z^{-2}}{2} - \frac{z^{-1}}{2} - \cdots. \] And now Corollary 8.27 gives \[ \int_{C[2,3]} \frac{\exp(z)}{\sin(z)} \, dz = -2\pi i e^z. \] Putting it all together yields the integral we've been after for two chapters: \[ \int_{C[1,2]} \frac{\exp(z)}{\sin(z)} \, dz = 2\pi i (1 - e^z). \] ## Exercises ### 8.1 For each of the following series, determine where the series converges absolutely and where it converges uniformly: (a) \[ \sum_{n=0}^{\infty} \frac{2^n}{(n+1)!} z^{2n} \] (b) \[ \sum_{n=0}^{\infty} \left(\frac{1}{(-3)^n}\right)^k \] ### 8.2 What functions are represented by the series in the previous exercise? ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 172 Context: 8.24. Prove: If \( f \) is entire and \( \text{Im}(f) \) is constant on the closed unit disk then \( f \) is constant. 8.25. (a) Find the Laurent series for \( \frac{1}{z^2} \) centered at \( z = 0 \). (b) Prove that \( f: \mathbb{C} \to \mathbb{C} \) is entire, where \[ f(z) = \begin{cases} \frac{1}{z} & \text{if } z \neq 0 \\ -1 & \text{if } z = 0 \end{cases} \] 8.26. Find the Laurent series for \( \sec(z) \) centered at the origin. 8.27. Suppose that \( f \) is holomorphic at \( z_0 \), \( f(z_0) = 0 \), and \( f'(z_0) \neq 0 \). Show that \( f \) has a zero of multiplicity 1 at \( z_0 \). 8.28. Find the multiplicities of the zeros of each of the following functions: (a) \( f(z) = e^{z} - 1 \), \( z_0 = 2k\pi i \), where \( k \) is any integer. (b) \( f(z) = \sin(z) - \tan(z) \), \( z_0 = 0 \). (c) \( f(z) = \cos(z) - 1 + z^2 \sin'(z_0) \), \( z_0 = 0 \). 8.29. Find the zeros of the following functions and determine their multiplicities: (a) \( 1 + z^2 \) (b) \( \sin(z) \) (c) \( z^2 \cos(z) \) 8.30. Prove that the series of the negative-index terms of a Laurent series \[ \sum_{k < 0} c_k (z - z_0)^k \] converges for \[ \frac{1}{R_1} < |z - z_0| < \frac{1}{R_0} \] for some \( R_1 \), and that the convergence is uniform in \( \{ z \in \mathbb{C} : |z - z_0| \geq r_1 \} \) for any fixed \( r_1 < R_1 \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 176 Context: # Isolated Singularities and the Residue Theorem **Definition.** If \( f \) is holomorphic in the punctured disk \( D(r, R) \) for some \( R > 0 \) but not at \( z = a \), then \( a \) is an isolated singularity of \( f \). The singularity \( a \) is called: 1. **(a) removable** if there exists a function \( g \) holomorphic in \( D(a, R) \) such that \( f(z) = g(z) \) for \( z \in D(a, R) \). 2. **(b) a pole** if \( \lim_{z \to a} |f(z)| = \infty \). 3. **(c) essential** if \( \lim_{z \to a} |f(z)| < \infty \) and it is neither removable nor a pole. ## Example 9.1 Let \( f: \mathbb{C} \setminus \{0\} \to \mathbb{C} \) be given by \( f(z) = \frac{e^{\frac{1}{z}}}{z} \). Since \[ \exp(z) - 1 = \sum_{n=1}^{\infty} \frac{1}{n!} z^n, \] the function \( g: \mathbb{C} \to \mathbb{C} \) defined by \[ g(z) = \sum_{n=0}^{\infty} \frac{1}{(k + 1)z^k}, \] which is entire (because this power series converges in \( \mathbb{C} \)), agrees with \( f \) on \( \mathbb{C} \setminus \{0\} \). Thus \( f \) has a removable singularity at \( 0 \). ## Example 9.2 In Example 8.23, we showed that \( f: \mathbb{C} \setminus \{ \infty \} \to \mathbb{C} \) given by \( f(z) = \frac{1}{z} \) has a removable singularity at \( \infty \), because we proved that \[ g : D(0, R) \text{ defined by } g(z) = \begin{cases} \frac{1}{z} & \text{if } z \neq 0 \\ 0 & \text{if } z = 0 \end{cases} \] is holomorphic in \( D(0, R) \) and agrees with \( f \) on \( D(0, R). \) ## Example 9.3 Let \( g: \mathbb{C} \setminus \{0\} \to \mathbb{C} \) given by \( f(z) = \frac{1}{z} \) has a pole at \( 0 \): \[ \lim_{z \to 0} |f(z)| = \infty. \] ## Example 9.4 The function \( f: \mathbb{C} \to \mathbb{C} \) given by \( f(z) = \exp\left(\frac{1}{z}\right) \) has an essential singularity at \( 0 \): the two limits \[ \lim_{z \to 0} \exp\left(\frac{1}{z}\right) = \infty \quad \text{and} \quad \lim_{z \to 0} \exp(z) = 0. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 178 Context: # Isolated Singularities and the Residue Theorem But this power series is holomorphic in \( D(z_0, r) \); so \( z_0 \) is a removable singularity. (b) Suppose that \( z_0 \) is a pole of \( f \). Since \( f(z) \to \infty \) as \( z \to z_0 \), we may assume that \( r \) is small enough that \( f(z) \) is holomorphic in \( D(z_0, r) \). Then \( g \) is holomorphic in \( D(z_0, r) \) and \[ \lim_{z \to z_0} (z - z_0)^n f(z) = 0, \] so part (a) implies that \( g \) has a removable singularity at \( z_0 \). More precisely, the function \( g: D(z_0, r) \to \mathbb{C} \) defined by \[ g(z) = \begin{cases} \frac{f(z)}{(z - z_0)^n} & \text{if } z \in D(z_0, r) \\ 0 & \text{if } z = z_0 \end{cases} \] is holomorphic. By Theorem 8.14, there exists a positive integer \( n \) and a holomorphic function \( h \) on \( D_{z_0, r} \) such that \( h(z_0) \neq 0 \) and \( g(z) = \frac{(z - z_0)^n}{h(z)} \). Actually, \( h(z) \) is holomorphic for all \( z \in D(z_0, r) \) since \( g(z) \) is holomorphic in \( D(z_0, r) \). Thus \[ \lim_{z \to z_0} (z - z_0)^n f(z) = \lim_{z \to z_0} \frac{(z - z_0)^n}{h(z)} \cdot s(z) = \frac{z - z_0}{h(z_0)} \lim_{z \to z_0} \frac{1}{h(z_0)} = 0. \] Note that \( h \) is holomorphic and non-zero on \( D(z_0, r) \), \( n > 0 \), and \[ f(z) = \frac{1}{s(z)} - \frac{1}{h(z)} \] for all \( z \in D(z_0, r) \). Conversely, suppose \( z_0 \) is not a removable singularity and \[ \lim_{z \to z_0} (z - z_0)^n f(z) = \infty \] for some non-negative integer \( n \). We choose the smallest such \( n \). By part (a), \( h(z) = (z - z_0)^{-n} f(z) \) has a removable singularity at \( z_0 \), so there is a holomorphic function \( g \) on \( D(z_0, r) \) that agrees with \( h \) on \( D(z_0, r) \). Now \( f(z) \) is not defined at \( z_0 \), since \( n \) was chosen as small as possible and \( n \) is a non-negative integer less than \( n \); we must have \[ g(z) = \lim_{z \to z_0} (z - z_0)^n f(z). \] Summarizing, \( g \) is holomorphic. #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 180 Context: ```markdown # Theorem 9.7 (Casorati–Weierstrass) If \( f \) is an essential singularity of \( r \) and \( r \) is any positive real number, then every \( u \in C \) is arbitrarily close to a point in \( f(B(z_0, r)) \). That is, for any \( x \in C \) and \( \epsilon > 0 \) there exists \( z \in B(z_0, r) \) such that \( |u - f(z)| < \epsilon \). In the language of topology, Theorem 9.7 says that the image of any punctured disk centered at an essential singularity is dense in \( C \). There is a stronger theorem, beyond the scope of this book, which implies the Casorati–Weierstrass Theorem 9.7. It is due to Charles Émile Picard (1856–1941) and says that the image of any punctured disk centered at an essential singularity misses at most one point of \( C \). (It is worth coming up with examples of functions that do not miss any point in \( C \) and functions that miss exactly one point. Try it!) ## Proof Suppose (by way of contradiction) that there exists \( w \in C \) and \( \epsilon > 0 \) such that for all \( z \in B(z_0, r) \) \[ |u - f(z)| \geq \epsilon. \] Then the function \( g(z) = \frac{f(z) - w}{z - z_0} \) stays bounded as \( z \to z_0 \) and so \[ \lim_{z \to z_0} (z - z_0) g(z) = \lim_{z \to z_0} \frac{f(z) - w}{z - z_0} = 0. \] (Proposition 9.5(a) tells us that \( f \) has a removable singularity at \( z_0 \).) Hence \[ \lim_{z \to z_0} \left| \frac{f(z) - w}{z - z_0} \right| = \infty \] and so the function \( f(z) - w \) has a pole at \( z_0 \). By Proposition 9.5(b), there is a positive integer \( n \) so that \[ \lim_{z \to z_0} (z - z_0)^{n+1} f(z) = \lim_{z \to z_0} (z - z_0)^n g(f(z)) = 0. \] Invoking Proposition 9.5 again, we conclude that the function \( f(z) \) has a pole or removable singularity at \( z_0 \), which implies the same holds for \( f(z) \), a contradiction. ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 186 Context: # 9.3 Argument Principle and Rouché's Theorem In the previous section, we saw how to compute integrals via residues, but in many applications, we actually do not have an explicit expression for a function that we need to integrate (or this expression is very complicated). However, it may still be possible to compute the value of a function at any given point. In this situation, we cannot immediately apply the Residue Theorem because we don't know where the singularities are. Of course, we could use numerical integration to compute integrals over any path, but computationally this task could be very resource intensive. But if we do know the singularities, we can compute the residues numerically by computing a finite number of the integrals over small circles around these singularities. And after that, we can apply the residue theorem to compute the integral over any closed path very effectively: we just sum up the residues inside this path. The argument principle that we study below, in particular, addresses this question. We start by introducing the logarithmic derivative. Suppose we have a differentiable function \( f \). Differentiating \( \log f \) (where \( \log \) is some branch of the logarithm) gives \( \frac{f'}{f} \), which is one good reason to call this quotient the logarithmic derivative of \( f \). It has some remarkable properties, one of which we would like to discuss here. Now let's say we have two functions \( f \) and \( g \) holomorphic in some region. Then the logarithmic derivative of their product behaves very nicely: \[ \frac{f g'}{g} = \frac{f' g + f g'}{g^2} \] We can apply this fact to the following situation: Suppose that \( f \) is holomorphic in a region \( G \) and has (finitely many) zeros \( z_1, z_2, \ldots, z_n \), of multiplicities \( m_1, \ldots, m_n \), respectively. By Theorem 8.14, we can express \( f \) as \[ f(z) = (z - z_1)^{m_1} (z - z_2)^{m_2} \cdots (z - z_n)^{m_n} g(z) \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 187 Context: # ARGUMENT PRINCIPLE AND ROUCHE'S THEOREM where \( g \) is also holomorphic in \( G \) and never zero. Let's compute the logarithmic derivative of \( f \) and play the same remarkable cancellation game as above: \[ f'(z) = \frac{f(z)}{g(z)} \] \[ n(z - z_j)^{-1}(z - z_j)^m g'(z) + (z - z_j)^m g(z)(z - z_j)^{-1} = \frac{m}{z - z_j} + \frac{n}{z - z_j} \frac{g'(z)}{g(z)}\tag{9.1} \] Something similar happens if \( f \) has finitely many poles in \( G \). In Exercise 9.19, we invite you to prove that, if \( p_1, \ldots, p_k \) are all the poles of \( f \) in \( G \) with order \( m_1, \ldots, m_k \), respectively, then the logarithmic derivative of \( f \) can be expressed as \[ \frac{f'(z)}{f(z)} = -\sum_{j=1}^{k} \frac{m_j}{z - p_j} - \sum_{j=1}^{k} \frac{m_j}{z - p_k} + \frac{g'(z)}{g(z)} \tag{9.2} \] where \( g \) is a function without poles in \( G \). Naturally, we can combine the expressions for zeros and poles, as we will do in a moment. ## Definition A function \( f \) is **meromorphic** in the region \( G \) if it is holomorphic in \( G \) except for poles. ### Theorem 9.17 (Argument Principle) Suppose \( f \) is meromorphic in a region \( G \) and \( \gamma \) is a positively oriented, simple, closed, piecewise smooth path that does not pass through any zero or pole of \( f \), and let \( \gamma^c \) denote \( Z(f; \gamma) \) the number of zeros of \( f \) inside \( \gamma \) counted according to multiplicity and \( P(f; \gamma) \) the number of poles of \( f \) inside \( \gamma \) counted according to multiplicity. Then \[ \frac{1}{2 \pi i} \int_{\gamma} \frac{f'(z)}{f(z)} \, dz = Z(\gamma) - P(f; \gamma) \] **Proof**. Suppose the zeros of \( f \) inside \( \gamma \) are \( z_1, \ldots, z_n \) of multiplicities \( m_1, \ldots, m_n \), respectively, and the poles inside \( \gamma \) are \( p_1, \ldots, p_k \) with order \( m_1, \ldots, m_k \), respectively. (You may meditate about the fact that there can be only finitely many zeros and poles inside \( \gamma \).) In fact, we may shrink \( \gamma \) if necessary, so that these are the only #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 188 Context: ```markdown 182 ISOLATED SINGULARITIES AND THE RESIDUE THEOREM zeros and poles in G. By (9.1) and (9.2), $$ f\left(z\right) = \frac{n_1}{z - z_1} + \frac{n_2}{z - z_2} - \frac{m_1}{z - p_1} + \frac{s'(z)}{g(z)} $$ where \(s\) is a function that is holomorphic in \(G\) (in particular, without poles) and never zero. Thanks to Cauchy’s Theorem 4.18 and Exercise 4.4, the integral is equal to $$ \int_{\gamma} \frac{f'(z)}{g(z)} dz = 2\pi i \left(n_1 + \ldots + n_j - m_1 - \ldots - m_k\right) \int_{\gamma} \frac{s'(z)}{g(z)}. $$ Finally, \( \frac{s'}{g} \) is holomorphic in \(G\) (because \(g\) is never zero in \(G\)), so Corollary 4.20 gives $$ \int_{\gamma} \frac{s'}{g} = 0. $$ As mentioned above, this beautiful theorem helps to locate poles and zeros of a function \(f\). The idea is simple: we can first numerically integrate \(f\) over a big circle \(\gamma\) that includes all possible paths over which we potentially will be integrating \(f\). Then the numerical value of \(\int_{\gamma} f\) will be close to an integer that, according to the Argument Principle, equals \(Z(f) - P(f)\). Then we can integrate \(\frac{f}{g}\) over a smaller closed path \(\gamma'\) that encompasses half of the interior of \(\gamma\) and \(Z(f) - P(f)\). Continuing this process for smaller and smaller regions will (after certain verification) produce small regions where \(f\) has exactly one zero or exactly one pole. Integrating \(f\) over the boundaries of those small regions that contain poles and dividing by \(Z(f)\) gives all residues of \(f\). Another nice relevant application of the Argument Principle is a famous theorem due to Rouche (1832–1910). **Theorem 9.18 (Rouche's Theorem):** Suppose \(f\) and \(g\) are holomorphic in a region \(G\) and \(r\) is a positively oriented, simple, closed, piecewise smooth path, such that \( |g(z)| < |f(z)| \) for all \( z \in \gamma \). Then $$ Z(f + g, \gamma) = Z(f, \gamma). $$ ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 191 Context: # Argument Principle and Rouché's Theorem ## 9.6 Suppose \( f \) has a simple pole (i.e., a pole of order 1) at \( z_0 \) and \( g \) is holomorphic at \( z_0 \). Prove that \[ \text{Res}_{z=z_0}(f(z)g(z)) = g(z_0) \text{Res}_{z=z_0}(f(z)). \] ## 9.7 Find the residue of each function at \( 0 \): (a) \( z^2 \cos(z) \) (b) \( \cos(z) \) (c) \( \frac{z^4 + 4z^2 + 5}{z^2 + z} \) (d) \( \exp(iz) - 1 \) ## 9.8 Use residues to evaluate the following integrals: (a) \( \int_{C(0,1)} \frac{dz}{z^4 + 4} \) (b) \( \int_{C(1)} \frac{dz}{z} \) (c) \( \int_{C(1)} \frac{dz}{z^2 + z - 2} \) (d) \( \int_{C(1)} \frac{\exp(z)}{(z + 2)^2} dz \) (e) \( \int_{C(1)} \frac{\exp(z)}{\sin(\cos(z))} dz \) ## 9.9 Use the Residue Theorem 9.10 to re-prove Cauchy's Integral Formulas (Theorems 4.27 & 5.1 and Corollary 8.11). ## 9.10 Revisiting Exercise 8.34, show that if \( f \) is even then \( \text{Res}_{z=z_0}(f(z)) = 0 \). ## 9.11 Suppose \( f \) has an isolated singularity at \( z_0 \). (a) Show that \( f' \) also has an isolated singularity at \( z_0 \). (b) Find \( \text{Res}_{z=z_0}(f') \). ## 9.12 Extend Proposition 9.14 by proving, if \( f \) and \( g \) are holomorphic at \( z_0 \), which is a double zero of \( g \), then \[ \text{Res}_{z=z_0}\left(\frac{f(z)}{g(z)}\right) = \frac{6f(z_0)g'(z_0)(f(z_0))^2 - 2f(z_0)f''(z_0)g(z_0)}{3(g'(z_0))^2}. \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 192 Context: 9.13. Compute \[ \int_{C(0, 2)} \frac{\cos(z)}{z^2 \sin^2(z)} dz. \] 9.14. Generalize Example 5.14 and Exercise 5.18 as follows: Let \( p(x) \) and \( q(x) \) be polynomials such that \( q(x) \neq 0 \) for \( x \in \mathbb{R} \) and the degree of \( q(x) \) is at least two larger than the degree of \( p(x) \). Prove that \[ \int_{-\infty}^{\infty} \frac{p(x)}{q(x)}\,dx \] equals \( 2\pi i \) times the sum of the residues of \( \frac{p(z)}{q(z)} \) at all poles in the upper half plane. 9.15. Compute \[ \int_{0}^{\infty} \frac{dx}{(1+x^2)}. \] 9.16. Generalize Exercise 5.19 by deriving conditions under which we can compute \[ \int_{-\infty}^{\infty} \frac{p(x)}{q'(x)}\,dx \] for polynomials \( p(x) \) and \( q(x) \), and give a formula for this integral along the lines of Exercise 9.14. 9.17. Compute \[ \int_{0}^{\infty} \frac{\cos(x)}{1+x^2} dx. \] 9.18. Suppose \( f \) is entire and \( a, b \in \mathbb{C} \) with \( a \neq b \) and \( |a|, |b| < r \). Evaluate \[ \int_{C(a, r)} \frac{f(z)}{(z-a)(z-b)} dz \] and use this to give an alternate proof of Liouville's Theorem 5.13. (Hint: Show that if \( f \) is bounded then the above integral goes to zero as \( r \) increases.) 9.19. Prove (9.2). 9.20. Suppose \( f \) is meromorphic in the region \( G \), \( f \) is holomorphic in \( G \), and \( \gamma \) is a positively oriented, simple, closed piecewise smooth path that does not pass through any zero or pole of \( f \), and \( \gamma' \). Denote the zeros and poles of \( f \) inside \( \gamma \) by \( z_1, z_2, \ldots, z_m \), respectively, counted according to multiplicity. Prove that \[ \frac{1}{2\pi i} \int_{\gamma} f(z) dz = \sum_{n=1}^{m} g(z_n) - \sum_{n=1}^{p} g(p_n). \] #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 194 Context: # Chapter 10 ## Discrete Applications of the Residue Theorem *All means (even continuous) sanctify the discrete end.* Doron Zeilberger On the surface, this chapter is just a collection of exercises. They are more involved than any of the ones we've given so far at the end of each chapter, which is one reason why we will lead you through each of the following ones step by step. On the other hand, these sections should really be thought of as a continuation of the book, just in a different format. All of the following problems are of a discrete mathematical nature, and we invite you to solve them using continuous methods—namely, complex integration. There are very few results in mathematics that so intimately combine discrete and continuous mathematics as does the Residue Theorem 9.10. ## 10.1 Infinite Sums In this exercise, we evaluate the sums \(\sum_{n=1}^{\infty} \frac{1}{n}\) and \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\). We hope the idea of how to compute such sums in general will become clear. 1. Consider the function \(f(z) = \frac{e^{z}}{z}\). Compute the residues at all the singularities of \(f\). (a) Let \(N\) be a positive integer and \(z_{N}\) be the rectangular path from \(N + iN\) to \(N + i(N - 1)\) to \(N - 1 + i(N - 1)\) to \(N - 1 + iN\) back to \(N + iN\). (b) Show that \(\lim_{N \to \infty} \int_{z_{N}} f(z) \, dz = 0\). 2. Use the Residue Theorem 9.10 to arrive at an identity for \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 196 Context: ``` Keeping \( x \) and \( y \) from (3), convince yourself that \[ \sum_{k=0}^{\infty} \left( \frac{2k}{k} \right) k^k = \frac{1}{2\pi i} \int \left( \frac{x^{n+1}}{z^{n+1}} \right) dz, \] use (3) to interchange summation and integral, and use the Residue Theorem 9.10 to evaluate the integral, giving an identity for \( \sum_{k=0}^{\infty} \left( \frac{1}{2} \right)^k \). ### 10.3 Fibonacci Numbers The Fibonacci\(^1\) numbers are a sequence of integers defined recursively through \[ f_0 = 0, \] \[ f_1 = 1, \] \[ f_n = f_{n-1} + f_{n-2} \quad \text{for } n \geq 2. \] Let \( F(z) = \sum_{n=0}^{\infty} f_n z^n \). 1. Show that \( F \) has a positive radius of convergence. 2. Show that the recurrence relation among the \( f_n \) implies that \( F(z) = \frac{z}{1 - z - z^2} \). (Hint: Write down the power series for \( F(z) \) and \( z^2 F(z) \) and rearrange both so that you can easily add.) 3. Verify that \[ \text{Res}_{z=0} \left( \frac{1}{z(1 - z - z^2)} \right) = f_0. \] 4. Use the Residue Theorem 9.10 to derive an identity for \( f_n \). (Hint: Integrate \[ \frac{1}{z(1 - z - z^2)} \] around \( C[0, R] \) and show that this integral vanishes as \( R \to \infty \). \(^1\) Named after Leonardo Pisano Fibonacci (1170–1250). ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 200 Context: # Discrete Applications of the Residue Theorem ## Historical Remark The sum in (10.1) is called a Dedekind* sum. It first appeared in the study of the Dedekind \( \eta \)-function \[ \eta(z) = \exp\left(\frac{\pi i}{12} z\right) \prod_{n=1}^{\infty} \left(1 - \exp(2 \pi i n z)\right) \] in the 1870s and has since intrigued mathematicians from such different areas as topology, number theory, and discrete geometry. The reciprocity law (10.2) is the most important and famous identity of the Dedekind sum. The proof that is outlined here is due to Hans Rademacher (1892–1969). --- *Named after Julius Wilhelm Richard Dedekind (1831–1916). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 201 Context: # Appendix: Theorems From Calculus First, it is necessary to study the facts, to multiply the number of observations, and then later to search for formulas that connect them so as to discern the particular law governing a certain class of phenomena. In general, it is not until after these particular laws have been established that one can expect to discover and articulate the more general law that connects them by bringing a multitude of apparently very diverse phenomena together under a single governing principle. _Augustin Louis Cauchy (1789–1857)_ Here we collect a few theorems from real calculus that we make use of in the course of the text. ## Theorem A.1 (Extreme-Value Theorem) Suppose \( K \subset \mathbb{R}^n \) is closed and bounded and \( f: K \to \mathbb{R} \) is continuous. Then \( f \) has a minimum and maximum value, i.e., \[ \min_{x \in K} f(x) \quad \text{and} \quad \max_{x \in K} f(x) \] exist in \( \mathbb{R} \). ## Theorem A.2 (Mean-Value Theorem) Suppose \( I \subset \mathbb{R} \) is an interval. \( f: I \to \mathbb{R} \) is differentiable, and \( x, x + h \in I \). Then there exists \( 0 < c < h \) such that \[ f(x + h) - f(x) = f'(c)(x + h - x). \] Many of the most important results of analysis concern combinations of limit operations. The most important of all calculus theorems combines differentiation and integration (in two ways): #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 203 Context: # Theorem A.9 (Leibniz's Rule) Suppose \( f \) is continuous on \([a, b] \times [c, d] \subset \mathbb{R}^2\) and the partial derivative \(\frac{\partial f}{\partial y}\) exists and is continuous on \([a, b] \times [c, d]\). Then \[ \frac{d}{dx} \int_c^d f(x, y) \, dy = \int_c^d \frac{\partial f}{\partial x}(x, y) \, dy. \] Leibniz's Rule follows from the Fundamental Theorem of Calculus (Theorem A.3). You can try to prove it, e.g., as follows: Define \( F(x) = \int_c^d f(x, y) \, dy \), get an expression for \( F(x) - F(a) \) as an iterated integral by writing \( F(x) - F(a) \) as the integral of \(\frac{\partial f}{\partial y}\), interchange the order of integrations, and then differentiate using Theorem A.3. # Theorem A.10 (Green's Theorem) Let \( C \) be a positively oriented, piecewise smooth, simple, closed path in \(\mathbb{R}^2\) and let \( D \) be the set bounded by \( C \). If \( f(x, y) \) and \( g(x, y) \) have continuous partial derivatives on an open region containing \( D \) then \[ \int_C f \, dx + g \, dy = \iint_D \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \, dx \, dy. \] # Theorem A.11 (L'Hôpital's Rule) Suppose \( I \subset \mathbb{R} \) is an open interval and either \( c \) is in \( I \) or \( c \) is an endpoint of \( I \). Suppose \( f \) and \( g \) are differentiable functions on \( I \setminus \{c\} \) with \( g(x) \) never zero. Suppose \[ \lim_{x \to c} f(x) = 0, \quad \lim_{x \to c} g(x) = 0, \quad \lim_{x \to c} \frac{f(x)}{g(x)} = L. \] Then \[ \lim_{x \to c} f(x) = L. \] There are many extensions of L'Hôpital's rule. In particular, the rule remains true if any of the following changes are made: - \( I \) is infinite. - \( I \) is unbounded and \( c \) is an infinite endpoint of \( I \). #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 204 Context: # Solutions to Selected Exercises Well here's another clue for you all. John Lennon & Paul McCartney ("Glass Onion," *The White Album*) ## 1.1 (a) \( 7 - i \) (b) \( 1 - i \) (c) \( -11 - 2i \) (d) \( -2 + 3i \) ## 1.2 (b) \( \frac{1}{3} - \frac{1}{2} i \) ## 1.3 (a) \( \sqrt{5}, -2 \) (b) \( 5\sqrt{5}, 5 - 10i \) (c) \( \frac{\sqrt{2 - 1} + \sqrt{(2 + \sqrt{9})}}{5} \) (d) \( 8 \) ## 1.4 (a) \( 2^{1/2} i \) (b) \( 2\sqrt{5}i^2 e^{2} \) (c) \( e^{5} \) ## 1.5 (a) \( -1 + i \) (b) \( 34i \) (c) \( -1 \) (d) \( 2 \) ## 1.9 \( e^{i \pi} = -1 \) ## 1.11 (a) \( x = e^{i k}, k = 0, 1, \ldots, 5 \) (b) \( x = 2 e^{i t^{5/4}}, k = 0, 1, 2, 3 \) ## 1.18 \( \cos \frac{\pi}{5} \) and \( \sin \frac{3\pi}{5} \) ## 1.33 For instance: (a) \( r(t) = (1 + i)t + e^{it}, 0 \leq t < 2\pi \). (b) \( r(t) = (2 - i)(1 - i)(2i) + (2)(1 + i + 3i), 0 \leq t \leq 1 \). (c) \( r(t) = 34e^{-i \pi} = 34e^{-i(t)}, 0 \leq t \leq \pi \). ## 2.2 (a) \( 1 + i \) ## 2.18 (a) differentiable and holomorphic in \( C \) with derivative \( e^{-z} e^{z} \) (b) nowhere differentiable or holomorphic (c) differentiable only on \( x + iy \in C : x = y \) with derivative \( 2x \), nowhere #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 206 Context: 200 # Solutions to Selected Exercises ## 4.1 (a) \(6\) (b) \(\pi\) (c) \(4\) (d) \(\sqrt{7} + \frac{1}{2} \sinh^{-1}(4)\) ## 4.5 (a) \(8i\) (b) \(0\) (c) \(0\) ## 4.6 (a) \(1-i\) (b) \(i\pi, -\pi, 0, 2\pi\) (c) \(i^2, -2, 0, 2r^2\) ## 4.7 (a) \(\frac{1}{3} (e^{3z} - e^{-3z})\) (b) \(0\) (c) \(\{g(\exp(3 + 3i)) - 1\}\) ## 4.12 \(\sqrt{3} i\) ## 4.18 (a) \(-4 + i\left(4 + \frac{1}{3}\right)\) (b) \((\ln(17) - 1) + i \left(\frac{\pi}{4} + 1\right)\) (c) \(2\sqrt{2} - 1 + 2i\sqrt{2}\) (d) \(\frac{1}{4} \sin(8) - 2 + i \left(2 - \frac{1}{4} \sinh(8)\right)\) ## 4.27 For \(|r| < |z| \Rightarrow 2\pi for \ |r| < |z|\) ## 4.30 \(\frac{24}{30}\) ## 4.34 \(0\) ## 4.35 For \(r = 1\): \( \frac{-1}{3} \); for \(r = 3\): \(0\); for \(r = 5\) ## 4.37 (a) \(2\pi\) (b) \(0\) (c) \(-\frac{3}{2}\) (d) \(\frac{2}{2r^2 - 1}\) ## 5.1 (a) \(2\pi\) (b) \(-6i\) (c) \(\pi\) (d) \(0\) ## 5.3 (a) \(0\) (b) \(2\pi i\) (c) \(0\) (d) \(\pi\) (e) \(f(0)\) ## 5.4 \(2\pi \exp{(iw)}\) ## 5.18 \(\frac{\sqrt{3}}{2}\) ## 5.20 \(\frac{\sqrt{z}}{\sin{\left(\frac{1}{\sqrt{2}} z\right)}}\) ## 7.1 (a) Divergent (b) Convergent (limit \(0\)) (c) Divergent (d) Convergent (limit \(2 -\frac{1}{2}\)) (e) Convergent (limit \(0\)) ## 7.3 \(\sum_{n=2}^{\infty} (-4)^n z^n = \sum_{n=3}^{\infty} \frac{1}{6} z^n = \frac{k + 1}{2} z^{n+2}\) #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 208 Context: # Index - **absolute convergence**, 129 - **absolute value**, 5 - **accumulation point**, 13, 24 - **addition**, 2 - **algebraically closed**, 103 - **alternating harmonic series**, 131 - **alternating zeta function**, 189 - **analytic**, 152 - **analytic continuation**, 158 - **antiderivative**, 76, 100, 196 - **Arg**, 59 - **arg**, 60 - **argument**, 5 - **axis** - imaginary, 5 - real, 5 - **bijection**, 31, 43 - **binary operation**, 2 - **binomial coefficient**, 189 - **boundary**, 13, 117 - **boundary point**, 13 - **bounded**, 13 - **branch of the logarithm**, 59 - **calculus**, 1, 195 - **Casorati–Weierstrass theorem**, 173 - **Cauchy's estimate**, 151 - **Cauchy's integral formula**, 85 - **extensions of**, 97, 151 - **Cauchy's theorem**, 81 - **Cauchy–Goursat theorem**, 82 - **Cauchy–Riemann equations**, 32 - **chain of segments**, 16 - **circle**, 12 - **closed** - disk, 13 - path, 16 - **set**, 13 - **coffee**, 88, 132, 173 - **comparison test**, 127 - **complete**, 123 - **complex number**, 2 - **complex plane**, 5 - extended, 47 - **complex projective line**, 47 - **composition**, 27 - **concatenation**, 74 - **conformal**, 30, 44, 118 - **conjugate**, 10 - **connected**, 14 - **continuous**, 26 - **contractible**, 83 - **convergence**, 122 - **pointwise**, 131 - **uniform**, 131 #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 209 Context: # Table of Contents 1. **Convergence** - Sequence - Series - Cosine - Cotangent - Cross Ratio - Curve - Cycloid 2. **Numerical Concepts** - Dedekind Sum - Dense - Derivative - Partial - Difference Quotient - Differentiable - Dilation - Discriminant 3. **Geometric Concepts** - Disk - Closed - Open - Punctured - Unit - Distance of Complex Numbers - Divergent - Domain - Double Series 4. **Mathematical Constants** - e - Embedding of R into C - Empty Set - Entire Functions - Essential Singularity - Euclidean Plane - Euler’s Formula 5. **Functions and Theorems** - Even - Exponential Function - Exponential Rules - Extended Complex Plane - Fibonacci Numbers - Field - Fixed Point - Frobenius Problem - Functions - Conformal - Even - Exponential - Logarithmic - Odd - Trigonometric - Fundamental Theorem - Of Algebra - Of Calculus 6. **Applications** - Geogebra - Geometric Interpretation of Multiplication - Geometric Series - Group 7. **Harmonic Analysis** - Harmonic - Harmonic Conjugate - Holomorphic - Homotopy - Hyperbolic Trig 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: 210 Context: ``` # Index - i, 4 - identity, 3 - identity map, 23 - identity principle, 154 - image - of a function, 27 - of a point, 23 - imaginary axis, 5 - imaginary part, 4 - improper integral, 104, 186 - infinity, 46 - inside, 88 - integral, 71 - path independent, 102 - integral test, 129 - integration by parts, 92 - interior point, 13 - inverse function, 31 - of a Möbius transformation, 43 - inverse parametrization, 74 - inversion, 45 - isolated point, 13 - isolated singularity, 170 - Jacobian, 65 - Jordan curve theorem, 88 - L'Hôpital's rule, 197 - Laplace, 110 - Laurent series, 157 - least upper bound, 124, 137 - Leibniz's rule, 83, 196 - length, 73 - limit - infinity, 46 - of a function, 24 - of a sequence, 122 - of a series, 126 - linear fractional transformation, 43 - Liouville's theorem, 104 - Log, 59 - log, 61 - logarithm, 59 - logarithmic derivative, 180 - max/min property for harmonic functions, 115, 155 - maximum - strong relative, 115 - weak relative, 116, 155 - mean-value theorem - for harmonic functions, 114 - for holomorphic functions, 86 - for real functions, 195 - meromorphic, 181 - minimum - strong relative, 115 - weak relative, 155 - Möbius transformation, 43 - modulus, 5 - monotone, 124 - monotone sequence property, 124 - Morera's theorem, 100 - multiplication, 2 - north pole, 51 ``` #################### File: A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf Page: 212 Context: ``` # Index - **simple**, 16 - **simply connected**, 101 - **sine**, 56 - **singularity**, 169 - **smooth**, 14 - **piecewise**, 72 - **south pole**, 51 - **stereographic projection**, 51 - **tangent**, 56 - **Taylor series expansion**, 148 - **topology**, 12, 88 - **translation**, 45 - **triangle inequality**, 11 - **reverse**, 11 - **Tribonacci numbers**, 191 - **trigonometric functions**, 56 - **trigonometric identities**, 7 - **trivial**, 26 - **uniform convergence**, 131 - **uniqueness theorem**, 154 - **unit circle**, 16 - **unit disk**, 16 - **unit sphere**, 51 - **vector**, 5 - **Weierstrass M-test**, 134 - **Weierstrass convergence theorem**, 163 ``` ########## """QUERY: Please summarize the whole context. It is important that you include a summary for each file. All files should be included, so please make sure to go through the entire context""" 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. 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Page 200, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 201, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 203, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 204, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 206, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 208, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 209, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 210, A%20First%20Course%20in%20Complex%20Analysis%20-%20Matthias%20Beck%2C%20Gerald%20Marchesi%2C%20Dennis%20Pixton%2C%20Lucas%20Sabalka%20%28PDF%29.pdf - Page 212 ================================================== **Elapsed Time: 96.51 seconds** ================================================== FINAL ANSWER Answer: ================================================== **Elapsed Time: 0.00 seconds** ==================================================