NotesAtHKU

The imaginary number $i=\sqrt{-1}$ . A complex number is usually denoted as $z=a+bi$ . In operations, the imaginary number can be considered as an unknown, though note that $i^2=-1$ (and so forth).

Useful items

Common properties

The modulus: $\|z\|=\sqrt{a^2+b^2}$

The conjugate: $conj(z) = \hat{z} = a-bi$

The argument: $arg(z) = \theta = 2tan^{-1}(\frac{b}{a+\|z\|})$

Polar form

We can write a complex number in polar form: $z=r(\cos\theta+i\sin\theta),\quad \text{where } r=\|z\|$ .

This represents the point on a circle with radius $r$ , at the angle $\theta$ .

Euler's formula

$re^{i\theta}=r(\cos\theta+i\sin\theta)$ .

Using the formula, we can easily derive:

$(\cos\theta+i\sin\theta)^n = (\cos{n\theta}+i\sin{n\theta})$
$(\cos\theta+i\sin\theta)^{-1} = (\cos{\theta}-i\sin{\theta})$
$(\sin\theta+i\cos\theta) = i(\cos{\theta}-i\sin{\theta})=ie^{-i\theta}$

Complex roots

Consider $nz=k$ , the solutions to the equation is:

$\underbrace{kw^0 + kw^1 + \dots + kw^{n-1}}_{n\text{ solutions}}\quad for\ w = e^(\frac{2\pi i}{n})$ ,

Set notation

Elements in a set are unique.

A=\{\underbrace{\underbrace{1}_{\text{An element}},2,3}_{\text{Elements}}\}\quad\quad B=\{\underbrace{x}_{\text{For all}}\ |\ \underbrace{0<x<2}_{\text{Such that}}\underbrace{,}_{\text{And}}\quad x\in\mathbb{R}\}=\{1\}

Image under complex function

Given a complex function $f(z)$ and a set of complex numbers $D=\{g(z)\in \mathbb{C}:h(z)\}$ , our goal is to solve for the function $f(D)$ in a form that we can identify the shape / image.

General steps

Find $f(D)$ by finding $f(g(z))$
Let $f(g(z))$ be $x'+y'i$ , then solve for $x$ and $y$ in terms of $x'$ and $y'$
(Note that $g(z)$ and $h(z)$ are functions of $x+yi$ )
Substitute your $x$ and $y$ to $f(D)$ , so that all unknowns are in terms of $x'$ and $y'$
Rearrange $h(z)$ to solve the shape of the image

Simple example

Let $D=\{x+iy \in \mathbb{C}\ \| \|x+iy\|>1\}$ , with complex function $f(z)=-1/z$ . Find $f(D)$ , then sketch the picture of $f(D)$ on the complex plane. 23 Dec, Part 2 Assignment 2 Question 3

We first find $f(D)$ by finding $f(x+yi)$ :

f(x+yi)=\frac{-1}{x+yi}\rightarrow f(D)=\{\frac{-1}{x+yi}\in \mathbb{C}||x+yi|>1\}

Then let $\frac{-1}{x+yi}$ be $x'+y'i$ , then solve for $x$ and $y$ in terms of $x'$ and $y'$

x'+y'i=\frac{-1}{x+yi}\rightarrow x+yi=\frac{-1}{x'+y'i}

Substitute the values we found into $f(D)$

f(D)=\{x'+y'i\in \mathbb{C}||\frac{-1}{x'+y'i}|>1\}

Then finally we rearrange the right side of the equation as:

\frac{1}{\sqrt{x'^2+y'^2}}>1

1>x'^2+y'^2

Therefore, we can conclude that the image is a filled circle with radius 1 (excluding edges):

FIX-ME: Insert a diagram here

A more elegant way

We can use Euler's formula to solve questions involving circles. Using the above example:

We let $x+yi$ as $re^{i\theta}$ and substitute:

D=\{re^{i\theta}\in\mathbb{C}|\underbrace{r>1}_{\text{As $\|z\|$ is $r$}}\}

Then following the usual steps:

f(D)=\{-r^{-1}e^{-i\theta}\in \mathbb{C}| r > 1\}

f(D)=\{r^{-1}\underbrace{e^{-i\theta+i\pi}}_{-1=e^{i\pi}}\in \mathbb{C}| r > 1\}

Let\ r'=r^{-1}, \theta'=\pi-\theta\rightarrow r=r'^{-1}

f(D)=\{r'e^{i\theta'}\in \mathbb{C}| r^{-1} > 1\}

f(D)=\{r'e^{i\theta'}\in \mathbb{C}| r < 1\}

Complex numbers