# Complex Analysis/Complex Functions/Complex Derivatives
## Complex differentiability

Let us now define what complex differentiability is.
Definition 2.3.1:
Let $S\subseteq \mathbb {C}$ , let $f:S\to \mathbb {C}$ be a function and let $z_{0}\in S$ . $f$ is called complex differentiable at $z_{0}$ if and only if there exists a $w\in \mathbb {C}$ such that:
    $\lim _{z\to z_{0} \atop z\in S}{\frac {f(z)-f(z_{0})}{z-z_{0}}}=w$

### Example 2.3.2

The function
    $f:\mathbb {C} \to \mathbb {C} ,f(z)={\bar {z}}$
is nowhere complex differentiable.

### Proof

Let $z_{0}\in \mathbb {C}$ be arbitrary. Assume that $f$ is complex differentiable at $z_{0}$ , i.e. that
    $\lim _{z\to z_{0} \atop z\in \mathbb {C} }{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}$
exists.
We choose
    ${\begin{aligned}A:=\{z\in \mathbb {C} |\Re z=\Re z_{0}\}\\B:=\{z\in \mathbb {C} |\Im z=\Im z_{0}\}\end{aligned}}$
Due to lemma 2.2.3, which is applicable since of course $\mathbb {C}$ is open, we have:
    $\lim _{z\to z_{0} \atop z\in A}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in \mathbb {C} }{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in B}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}$
But
    ${\begin{aligned}&\lim _{z\to z_{0} \atop z\in A}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in A}{\frac {\Re (z-z_{0})-i\Im (z-z_{0})}{\Re (z-z_{0})+i\Im (z-z_{0})}}=-1\\\\&\lim _{z\to z_{0} \atop z\in B}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in B}{\frac {\Re (z-z_{0})-i\Im (z-z_{0})}{\Re (z-z_{0})+i\Im (z-z_{0})}}=1\end{aligned}}$
a contradiction.$\Box$

## The Cauchy–Riemann equations

We can define a natural bijective function from $\mathbb {C}$ to $\mathbb {R} ^{2}$ as follows:
    $\Phi (x+yi):=(x,y)$
In fact, $\Phi$ is a vector space isomorphism between $\mathbb {C} ^{1}$ and $\mathbb {R} ^{2}$ .
The inverse of $\Phi$ is given by
    $\Phi ^{-1}:\mathbb {R} ^{2}\to \mathbb {C} ,\Phi ^{-1}(x,y)=x+yi$
Theorem and definitions 2.3.3:
Let $O\subseteq \mathbb {C}$ be open, let $f:O\to \mathbb {C}$ be a function and let $z_{0}=x_{0}+y_{0}i\in O$ . If $f$ is complex differentiable at $z_{0}$ , then the functions
    ${\begin{aligned}&u:\Phi (O)\to \mathbb {R} ,u(x,y)=\Re f(x+yi)\\&v:\Phi (O)\to \mathbb {R} ,v(x,y)=\Im f(x+yi)\end{aligned}}$
are well-defined, differentiable at $(x_{0},y_{0})$ and satisfy the equations
    ${\begin{aligned}&\partial _{x}u(x_{0},y_{0})=\partial _{y}v(x_{0},y_{0})\\&\partial _{y}u(x_{0},y_{0})=-\partial _{x}v(x_{0},y_{0})\end{aligned}}$
These equations are called the Cauchy-Riemann equations.

### Proof

1. We prove well-definedness of $u,v$ .
Let $(x,y)\in \Phi (O)$ . We apply the inverse function on both sides to obtain:
    $x+yi\in \Phi ^{-1}(\Phi (O))=O$
where the last equality holds since $\Phi$ is bijective (for any bijective $f:S_{1}\to S_{2}$ we have $f^{-1}{\biglf(S_{3}){\bigr )}=f{\biglf^{-1}(S_{3}){\bigr )}=S_{3}$ if $S_{3}\subseteq S_{1}$ ; see exercise 1).
3. We prove differentiability of $u$ and $v$ and the Cauchy-Riemann equations.
We define
    ${\begin{aligned}S_{1}:=\{z\in \mathbb {C} :\Re (z)=\Re (z_{0})\}\cap O\\S_{2}:=\{z\in \mathbb {C} :\Im (z)=\Im (z_{0})\}\cap O\end{aligned}}$
Then we have:
    ${\begin{aligned}\partial _{x}u(x_{0},y_{0})&=\lim _{x\to x_{0}}{\frac {u(x,y_{0})-u(x_{0},y_{0})}{x-x_{0}}}&\\&=\lim _{x\to x_{0}}{\frac {\Re {\biglf(x+y_{0}i){\bigr )}-\Re {\biglf(x_{0}+y_{0}i){\bigr )}}{x-x_{0}}}&\\&=\Re \left(\lim _{x\to x_{0}}{\frac {f(x+y_{0}i)-f(x_{0}+y_{0}i)}{x-x_{0}}}\right)&{\text{continuity of }}\Re \\&=\Re \left(\lim _{z\to z_{0} \atop z\in S_{2}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\right)&\\&=\Re \left(\lim _{z\to z_{0} \atop z\in S_{1}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\right)&{\text{lemma 2.2.3}}\\&=\Re \left(\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{yi-y_{0}i}}\right)&\\&=\Re \left((-i)\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{y-y_{0}}}\right)&i^{-1}=-i\\&=\Im \left(\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{y-y_{0}}}\right)&\\&=\partial _{y}v(x_{0},y_{0})\end{aligned}}$
From these equations follows the existence of $\partial _{x}u(x_{0},y_{0}),\partial _{y}v(x_{0},y_{0})$ , since for example
    $\lim _{z\to z_{0} \atop z\in S_{2}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}$
exists due to lemma 2.2.3.
The proof for
    $\partial _{y}u(x_{0},y_{0})=-\partial _{x}v(x_{0},y_{0})$
and the existence of $\partial _{y}u(x_{0},y_{0}),\partial _{x}v(x_{0},y_{0})$ we leave for exercise 2.$\Box$

## Holomorphic functions

Definitions 2.3.4:
Let $S\subseteq \mathbb {C}$ and let $f:S\to \mathbb {C}$ be a function. We call $f$ holomorphic if and only if for all $z_{0}\in S$ , $f$ is differentiable at $z_{0}$ . In this case, the function
    $f':S\to \mathbb {C} ,f'(z_{0})=\lim _{z\to z_{0} \atop z\in S}{\frac {f(z)-f(z_{0})}{z-z_{0}}}$
is called the complex derivative of $f$. We write $H(S)$ for the set of holomorphic functions defined on $S$ .

## Exercises

* Let $S_{1},S_{2},S_{3}$ be sets such that $S_{3}\subseteq S_{1}$ , and let $f:S_{1}\to S_{2}$ be a bijective function. Prove that $f^{-1}{\biglf(S_{3}){\bigr )}=f{\biglf^{-1}(S_{3}){\bigr )}=S_{3}$ .
* Let $O\subseteq \mathbb {C}$ be open, let $f:O\to \mathbb {C}$ be a function and let $z_{0}=x_{0}+y_{0}i\in O$ . Prove that if $f$ is complex differentiable at $z_{0}$ , then $\partial _{y}u(x_{0},y_{0})$ and $\partial _{x}v(x_{0},y_{0})$ exist and satisfy the equation $\partial _{y}u(x_{0},y_{0})=-\partial _{x}v(x_{0},y_{0})$ .
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