Lecture 18 - Complex Analysis

Overview of Complex Analysis of the Course

• Will do analysis using complex numbers.
• Concrete Goal: compute integrals such as ${\int }_{-\infty }^{+\infty }\frac{e^{i\omega t}}{t^2+a^2}dt=\frac{\pi }{a}{e}^{-a\mid \omega \mid }$ ($a>0$) using complex techniques.

Complex Numbers

The definition of a Complex Numbers as defined in the lecture.

All algebraic properties follow from the definition. For instance
$(x_1+i{y}_1)(x_2+i{y}_2)=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+y_1{x}_2).$
This operation is commutative, that is $z_1{z}_2=z_2{z}_1$.

Proposition

Let $z=x+iy$, and it should be non-zero. Then there exists another complex number $z^{-1}\in \mathbb{C}$ such that $z{z}^{-1}=1$. It is given by
$z^{-1}=\frac{x-iy}{x^2+y^2}$

Proof

We compute
$(x+iy)(x-iy)=x^2+y^2.$
This is non-zero. Then
$z{z}^{-1}=(x+iy)\frac{x-iy}{x^2+y^2}=1$

Remark

This means that $\mathbb{C}$ is a field like $\mathbb{R}$.

Definition - Absolute Value

Let $z=x+iy$. Its absolute value (or norm) is defined by
$\mid z\mid \sqrt{x^2+y^2}.$
An argument for $z$ is a real number $\varphi \in \mathbb{R}$ such that
$x=\mid z\mid \cos \varphi ,y=\mid z\mid \sin \varphi .$
This allows us to identify complex numbers with points in the plane ${\mathbb{R}}^2$:

(note that angles are only defined up to multiples of $2\pi$).

You could multiply pairs in ${\mathbb{R}}^2$ by
$(x_1,y_1)\times (x_2,y_2)=(x_1{x}_2,y_1{y}_2).$
But this does not correspond to the multiplication of complex numbers.

Polar Form

• Using the previous definitions, we can write any $z\in \mathbb{C}$ as:
  $z=x+iy=\mid z\mid (\cos \varphi +i\sin \varphi ).$

Definition

We introduce the following notations:
$r:=\mid z\mid ,$
$e^{i\varphi }:=\cos \varphi +i\sin \varphi .$
Then we have that $z=r{e}^{i\varphi }$, which we call the polar form of $z$.

Remark

For now, $e^{i\varphi }$ is just a symbol. Later we will identify it with the Complex Exponential Function

We have that $e^{i\varphi }$ shares many properties with $e^x$.

Proposition

Let $\varphi ,\theta \in \mathbb{R}$. Then:

1. $e^{i\varphi }{e}^{i\theta }=e^{i(\varphi +\theta )}$,
2. $e^{i0}=1$,
3. $\frac{1}{e^{i\varphi }}=e^{-i\varphi }$,
4. $e^{i(\varphi +2\pi k)}=e^{i\varphi }$, for $k\in \mathbb{Z}$,
5. $\mid e^{i\varphi }\mid =1$,
6. $\frac{d{e}^{i\varphi }}{d\varphi }=i{e}^{i\varphi }$.

Proof (only for 3.)

According to a previous result, we have:
$(\cos \varphi +i\sin \varphi {)}^{-1}=\frac{\cos \varphi -i\sin \varphi }{\stackrel{=1}{\overbrace{(\cos \varphi {)}^2+(\sin \varphi {)}^2}}}$
$=\cos \varphi -i\sin \varphi$
$=e^{-i\varphi },$
using $\cos (-\varphi )=\cos \varphi$ and $\sin (-\varphi )=-\sin \varphi$.

Complex Conjugation

Definition of Complex Conjugation, as defined in the lecture.

Triangle Inequality

Definition of Triangle Inequality, as defined in the lecture.

But we want to show a similar result for $\mathbb{C}$.

Lemma

Let $z\in \mathbb{C}$, then we can take a look at the $\mathrm{Re}$ and $\mathrm{Im}$:
$-\mid z\mid \le Rez\le \mid z\mid$
$-\mid z\mid \le Imz\le \mid z\mid .$

Proof

Recall that for $z=x+iy$ we have $\mid z\mid =\sqrt{x^2+y^2}$. Then the results follow from the following inequalities ($a,b\in \mathbb{R}$):
$-\sqrt{a^2+b^2}\le -\sqrt{a^2}\le \mid a\mid \le \sqrt{e^2}\le \sqrt{a^2+b^2}.$
This gives the result.

We use this to prove the triangle inequality for $\mathbb{C}$.

Proposition

Let $z_1,z_2\in \mathbb{C}$ we have
$\mid z_1+z_2\mid \le \mid z_1\mid +\mid z_2\mid .$

Proof

We compute:
$\mid z_1+z_2{\mid }^2=(z_1+z_2)(\overline{z_1}+\overline{z_2})$
$=z_1\overline{z_1}+z_1\overline{z_2}+z_2\overline{z_1}+z_2\overline{z_2}=\mid z_1{\mid }^2+2\mathrm{Re}(z_1\overline{z_2})\mid z_2{\mid }^2$
(Noting that $z_2\overline{z_1}=\overline{z_1\overline{z_2}}$).

Next we use the property that $\mathrm{Re}z\le \mid z\mid$. Then
$\mid z_1+z_2{\mid }^2\le \mid z_1{\mid }^2+2\mid z_1\overline{z_2}\mid +\mid z_2{\mid }^2$
$=\mid z_1{\mid }^2+2\mid z_1\mid \times \mid \overline{z_2}\mid +\mid z_2{\mid }^2$
$=\mid z_1{\mid }^2+2\mid z_1\mid \mid z_2\mid +\mid z_2{\mid }^2$
$=(\mid z_1\mid +\mid z_2\mid {)}^2.$
Taking square roots of both (non-negative) sides, we get the result.

This means that $\mathbb{C}$ is a normed space.

Limits

Before being able to define a limit, we need to define a complex function, from the lecture.

Definition

Complex Limits as defined in the lecture

Proposition

Let $f$ and $g$ be complex functions.

Suppose that ${\lim }_{z\to z_0}f(z)$ and ${\lim }_{z\to z_0}g(z)$ exist. Then:

1. For $a,b\in \mathbb{C}$ we have ${\lim }_{z\to z_0}(af(z)+bg(z))=a{\lim }_{z\to z_0}f(z)+b{\lim }_{z\to z_0}g(z)$,
2. We have ${\lim }_{z\to z_0}(f(z)\times g(z))=\left({\lim }_{z\to z_0}f(z)\right)\times \left({\lim }_{z\to z_0}g(z)\right)$.

Example

(Computing the limit, but will not happen again, just for understanding)

Let $f(z)=z^2$. We want to show that ${\lim }_{z\to 2}f(z)=4$. Fix $\epsilon >0$ and observe that
$f(z)-4=z^2-4=(z-2{)}^2+4(z-2).$
Write $\mid z-2\mid <\delta$ for some $\delta$. Then using the triangle inequality, we get
$\mid f(z)-4\mid =\mid (z-2{)}^2+4(z-2)\mid$
$\le \mid z-2{\mid }^2+4\mid z-2\mid$
$<{\delta }^2+4\delta .$
It suffices to pick $\delta$ such that ${\delta }^2+4\delta \le \epsilon$ (which is possible).

A much easier way to compute the following:
First ${\lim }_{z\to z_0}z=z_0$, then
${\lim }_{z\to 2}{z}^2=\left({\lim }_{z\to 2}z\right)\times \left({\lim }_{z\to 2}z\right)=2\times 2=4$