# Complex Analysis

## Peter Zenz - Fall 2022

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### Lecture 1 - Introduction, Complex Numbers, and Polar Coordinates

#### Introduction and Overview

#### Complex Definitions

__Addition__: \((a + bi) + (c + di) = (a + c) + (b + d)i\)__Multiplication__: \( (a + bi)(c + di) = ac - bd + i(da + bc) \)

__Modulus/Absolute Value__: \( |z| = |x + yi| := \sqrt{x^2 + y^2} \)

__Division__: For \(z = x + yi \neq 0\), we want an \(a + bi\) such that $$ (x + yi)(a + bi) = 1. $$ \(a + bi = \frac{1}{x + yi} \cdot \frac{x - yi}{x - yi} = \frac{x-yi}{x^2 + y^2}\), so \(a=\frac{x}{x^2+y^2}\) and \(b=\frac{-y}{x^2+y^2}\).__Complex Conjugation__: \( \overline{x + yi} = x - yi \)- This conjugation has some nice properties:
- \(\bar{\bar{z}} = z\)
- \(\overline{z+w} = \overline{z} + \overline{w}\)
- \(\overline{zw} = \overline{z} \cdot \overline{w} \)
- \(|z| = |\overline{z}|\)
- \( |z|^2 = z\overline{z} \)
- \( Re(z) = \frac{z + \overline{z}}{z} \;\; ; \;\;\ Im(z) = \frac{z - \overline{z}}{z} \)
- \( |zw| = |z||w| \)

#### Polar Coordinates

*argument*of \(z\). Because this \(\theta\) is brought from a multivalued function, we want to denote a

*principal value*of \(arg(z)\), denoted by \(Arg(z)\), as \(\theta\) restricted between \(-\pi < \theta \leq \pi \).

### Lecture 2 — Polar Coordinates and Complex Exponentials

#### Complex Multiplication Geometrically

**multiplying their lengths and adding their angles**.

#### Facts from Real Analysis

- For every \( r \in \mathbb{R}, \,\, \displaystyle\sum_{n=0}^{\infty}\frac{r^n}{n!} \) converges.
- If \( a_n \in \mathbb{C} \) and \( \displaystyle\sum_{n=0}^{\infty}|a_n| \) converges, then \( \displaystyle\sum_{n=0}^{\infty} a_n \) converges.
- If \( \displaystyle\sum_{n=0}^{\infty} a_n \) and \( \displaystyle\sum_{n=0}^{\infty} b_n \) converge absolutely, then \( \displaystyle\sum_{n=0}^{\infty}\displaystyle\sum_{k=0}^{n}a_nb_{n-k}\) converges to \( \displaystyle\sum_{n=0}^{\infty}a_n\displaystyle\sum_{m=0}^{\infty}b_m \).

#### Complex Exponentials

### Lecture 3 — Complex Exponentials, Logarithms, and Power Functions

__Properties__:

- \( |e^{i\theta}| = 1 = \sqrt{\cos^2\theta + \sin^2\theta} \)
- \( e^{\overline{i\theta}} = e^{-i\theta} \)
- \( \frac{1}{e^{i\theta}} = e^{-i\theta} \)

**horizontal lines to rays**and

**vertical lines to circles**.

#### Proving Trig Identities From Exponentials

#### Complex Logarithms

We want to invert \( e^z \). There's a problem: \( e^z \) is not injective. (Compare this relationship with \(\sin x\) and \(\arcsin x\) restricted to \([-1, 1]\).) Here's how we can do it:

For \(z \neq 0 \in \mathbb{C}\), we define $$\begin{align} \log z &= \log |z| + arg(z) \\ &= \log |z| + iArg(z) + 2\pi ik. \end{align}$$

Values of \(\log z\) are complex numbers such that \(e^w = z\) $$\begin{align} e^2 = e^{\log z} &= e^{\log |z|}e^{iArg(z)}e^{2\pi ik}\\ &= |z| \cdot e^{iArg(z)} \cdot 1 = z. \end{align}$$

On the other hand, suppose \(w = u + iv\) such that \(e^2 = z = re^{i\theta}\). Then, $$\begin{align} r = |z| = e^u \Rightarrow u = \log |z|\;\; \text{ and }\;\; v = arg (z). \end{align}$$

The principal value of \( \log z \) is defined by \( \text{Log }(z) = \log |z| + iArg(z). \)

Note: \(\log z\) is a multivalued function, but \(\text{Log } z\) is a single-valued function with values in the horizontal strip \(-\pi < Im(z) \leq \pi\).

Compute \(\text{Log } i\): $$\begin{align} \text{Log } i &= \log |i| + iArg(i) \\ &= log(1) + i\frac{\pi}{2} &= \frac{i\pi}{2} \end{align}$$

#### Power Functions

For \(z,\,\alpha \in \mathbb{C}\) where \(z \neq 0\), we can define a general power function as $$ z^\alpha := e^{\alpha\log z}. $$ So, $$ e^{\alpha(\log |z| + iArg(z) + 2\pi ik)}. $$

So we can now use exponents with complex numbers more generally. But we should be careful!
Because things don't behave exactly like the reals here. **The power rule of distribution
doesn't hold**, i.e.
$$ z^\alpha z^\beta \neq z^{\alpha + \beta}. $$

### Lecture 4 —

#### Complex Roots

We want some \( w \in \mathbb{C} \) such that \( w^2 = z \). We'll write \(z = re^{i\theta}\) and \( w = se^{i\phi} \). Then, $$ w^2 = s^2e^{i2\phi} = re^{i\theta}. $$ This means that $$ s^2 = r \;\;\;\; \text{and} \;\;\;\; 2\phi = \theta. $$ So we conclude that \( w = \sqrt{r}e^{i\theta / 2} \). But there's another solution: \( w = \sqrt{r}e^{i(\frac{\theta}{2} + \pi)} \). Note that there are two solutions.

**In general, an \( n^{\text{th}} \) root has \(n\) solutions in \(\mathbb{C}\).**

__General \(n^{\text{th}}\) Root__

We want \(w\in\mathbb{C}\) where \(w^n = z\). Well we
have
$$\begin{align} \Leftrightarrow& s^ne^{in\phi} = re^{i\theta}\\ \Leftrightarrow& s^n = r
\;\;\;\;\text{and}\;\;\;\;n\phi = \theta + 2\pi k\end{align}$$
So, \(s = \sqrt[n]{r}\) and \( \phi = \frac{\theta}{n} + \frac{2\pi k}{n} \) with \( k =
0,\dots,n-1 \).

Compute the square root of \( z = -1 - i \).

Well, \( -1 - i = \sqrt{2}e^{i\frac{5\pi}{4}}
\). We get \(w_1 = \sqrt[4]{2}e^{i\frac{5\pi}{8}}\) and \( w_2 = \sqrt[4]{2}e^{i\frac{13\pi}{8}}
\).

#### Complex Trigonometric Functions