---
lecture: 18
date: 2025-03-20
---
# Overview of Complex Analysis of the Course
- Will do **analysis** using complex numbers.
- **Concrete Goal:** compute integrals such as $\int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{t^2 + a^2} \, dt = \frac{\pi}{a}e^{-a\mid \omega \mid}$ ($a \gt 0$) 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
$$(x_{1}+iy_{1})(x_{2}+iy_{2}) = (x_{1}x_{2} -y_{1}y_{2}) + i(x_{1}y_{2} + y_{1}x_{2}).$$
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 $zz^{-1} = 1$. 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
$$zz^{-1} = (x+iy) \frac{x-iy}{x^2 + y^2} = 1$$
> [!info] Remark
> This means that $\mathbb{C}$ is a **field** like $\mathbb{R}$.

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

![[Drawing 2025-03-20 10.50.15.excalidraw.dark.svg]]
%%[[Drawing 2025-03-20 10.50.15.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-20 10.50.15.excalidraw.light.svg|light exported image]]%%
(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:
$$z = x + iy = \mid z \mid (\cos \phi + i \sin \phi).$$
## Definition
We introduce the following notations:
$$r := \mid z \mid,$$
$$e^{i\phi} := \cos \phi + i \sin \phi.$$
Then we have that $z = r e^{i\phi}$, which we call the **polar form** of $z$.
> [!info] Remark
> For now, $e^{i\phi}$ is just a symbol. Later we will identify it with the [[Complex Exponential Function]]

We have that $e^{i\phi}$ shares many properties with $e^{x}$.
## Proposition
Let $\phi, \; \theta \in \mathbb{R}$. Then:
1. $e^{i \phi} e^{i \theta} = e^{i (\phi + \theta)}$,
2. $e^{i 0} = 1$,
3. $\frac{1}{e^{i \phi}} = e^{-i \phi}$,
4. $e^{i(\phi + 2 \pi k)} = e^{i \phi}$, for $k \in \mathbb{Z}$,
5. $\mid e^{i \phi} = 1$,
6. $\frac{d e^{i \phi}}{d \phi} = i e^{i \phi}$.
### Proof (only for 3.)
According to a previous result, we have:
$$(\cos \phi + i \sin \phi)^{-1} = \frac{\cos \phi - i \sin \phi}{\overbrace{(\cos \phi)^2 + (\sin \phi)^2}^{=1}}$$
$$= \cos \phi - i \sin \phi$$
$$= e^{- i \phi},$$
using $\cos(-\phi) = \cos \phi$ and $\sin(-\phi) = -\sin \phi$.
# Complex Conjugation
Definition [[Complex Conjugation]] as defined in the lecture.