# Cauchy's integral formula
## Introduction

The Cauchy integral formula, alongside the Cauchy's integral theorem, is one of the central statements in complex analysis. Here, we present two variants: the 'classical' formula for circular disks and a relatively general version for null-homologous Chain. Note that we will deduce the circular disk version from Cauchy's integral theorem, but for the general variant, we proceed in the opposite direction.

## For Circular Disks
### Statement

Let $G\subseteq \mathbb {C}$ be an open set, $D$ a circular disk with ${\bar {D}}\subseteq G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then, we have
    $f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw$
for each $z\in D$.

### Proof 1

By slightly enlarging the radius of the circular disk, we find an open circular disk $U$ such that ${\bar {D}}\subseteq U\subseteq G$. Define $g\colon U\to \mathbb {C}$ by
    $g(w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.$

### Proof 2

The function $g$ is continuous on $U$ and holomorphic on $U-{z}$. Thus, we can apply the Cauchy integral theorem on $U$ and obtain
    $0=\int _{\partial D}g(w),dw=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}$
For $z\in D$, define $h(z):=\int _{\partial D}{\frac {dw}{w-z}}$. Then $h$ is holomorphic with
    $h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}$

### Proof 3

Since the integrand ${\frac {dw}{(w-z)^{2}}}$ has a primitive in $D$, we find
    $h'(z)=-\int _{\partial D}{\frac {dw}{(w-z)^{2}}}=0$

### Proof 4

Because $h'(z)=0$ throughout $D$, it follows that $h$ is constant. Thus, $h(z)$ always takes the same value as at the center $h(z_{0})$ of the disk $D$, i.e., $h(z_{0})=2\pi i$. Hence,
    $0=\int _{\partial D}{\frac {f(w)}{w-z}},dw-f(z)\int _{\partial D}{\frac {dw}{w-z}}\iff f(z)={\frac {1}{2\pi i}}\int _{\partial D}{\frac {f(w)}{w-z}},dw$
This proves the statement.

## For Cycles in Arbitrary Open Sets
### Statement

Let $G\subseteq \mathbb {C}$ be an open set, $\Gamma$ a null-homologous cycle in $G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then,
    $n(\Gamma ,z)\cdot f(z)={\frac {1}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{w-z}},dw$
for each $z\in G\setminus \mathrm {spur} (\Gamma )$, where $n(\Gamma ,\cdot )$ denotes the winding number.

### Proof 1

Define a function $g\colon G^{2}\to \mathbb {C}$ by
    $g(z,w):=\left\{{\begin{array}{ll}{\frac {f(w)-f(z)}{w-z}}&w\neq z\\f'(z)&w=z\end{array}}\right.$
defined.

### Proof 2: continuous

We demonstrate the continuity in both variables. Let $(w_{0},z_{0})\in U\times U$ with $z_{0}\neq w_{0}$, then $g$ is given in the vicinity of $(w_{0},z_{0})$ by the above formula and is trivially continuous.Now let $z_{0}=w_{0}$. We choose a $\delta$-neighborhood $U_{\delta }(z_{0})\subset \subset G$ and examine $g(w,z)-g(z_{0},z_{0})$ auf $U_{\delta }(z_{0})\times U_{\delta }(z_{0})$ . a) In the case $w=z$: :
    $g(z,z)-g(z_{0},z_{0})=f'(z)-f'(z_{0})$

### Proof 3

b) In the case $w\neq z$:
    $g(w,z)-g(z_{0},z_{0})={\frac {f(w)-f(z)}{w-z}}-f'(z_{0})={\frac {1}{w-z}}\int _{[z,w]}(f'(v)-f'(z_{0}))dv$
Now, as a consequence of Cauchy's formulas for circles! the derivative $f'$ is continuous in $z_{0}$. For a given $\epsilon >0$ we can choose $\delta >0$ such that
    $|f'(v)-f'(z_{0})|<\epsilon$
for all $v\in U_{\delta }(z_{0})$.

### Proof 4

This implies, in case a:
    $|g(z,z)-g(z_{0},z_{0})|<\epsilon ;$
and in case b:
    $|g(w,z)-g(z_{0},z_{0})|\leq {\frac {1}{|w-z|}}|w-z|\cdot \sup \limits _{w\in [w,z]}|f'(v)-f'(z_{0})|<\epsilon .$
We now define
    $h_{0}(z)=\int _{\Gamma }g(w,z)dw.$
$h_{0}$ function is continuous on whole of $G$ ; we will show that it is even holomorphic. For this, we use Morera's theorem.

### Proof 5

Let $\gamma$ be the oriented boundary of a triangle that lies entirely with in $G$ . We must show
    $\int _{\gamma }h_{0}(z)dz=0$
prove it is
    $\int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.$
because the integrations are commutable due to the continuity of the integrand on $G\times G$ For fixed , $w$ the function is $g(w,z)$ in the Variable $z$ continuous in and holomorphic for $w\neq z$, hence holomorphic everywhere.

### Proof 6

By Goursat's theorem, it follows that
    $\int _{\gamma }g(w,z)dz=0.$
this of course also mean that
    $\int _{\gamma }h_{0}(z)dz=\int _{\Gamma }\int _{\gamma }g(w,z)dz\,dw=0.$
so far we have not yet exploited the conditions above $\Gamma$. We will do so
    $G_{0}=\{z\in \mathbb {C} :n(\Gamma ,z)=0\}$.

### Proof 7

Since on $G\cap G_{0}$ the function $h_{0}$ has a simpler form, namely
    $h_{0}(z)=\int _{\Gamma }{\frac {f(w)}{w-z}}dw=h_{1}(z),$
and since the function $h_{1}$ is clearly holomorphic on the entire $G_{0}$, we can extend $h_{0}$ to a holomorphic function $h$ defined on the entire by$G\cup G_{0}$
    $h(z)=\left\{{\begin{array}{ll}h_{0}(z)&z\in G\\h_{1}(z)&z\in G_{0}\end{array}}\right.$
Now $\Gamma$ is null-homologous in , and thus
    $G\cup G_{0}=\mathbb {C} ,$
i.e. $h$ is an entire function.

### Proof 8

For $h$ we have$G_{0}$ the notation:
    $|h(z)|=|h_{1}(z)|\leq {\frac {1}{dist(z,\Gamma )}}L(\Gamma )\max _{\Gamma }|f|\;\;\;(*);$
where $L(\Gamma )=\sum |n_{k}|L(\gamma _{k})$, if $\Gamma =n_{k}\cdot \gamma _{k}$ is.
$G_{0}$ contains the complement of a sufficiently large circle around 0. Therefore, the above inequality holds for all $z$ in this region: it follows that $h$ is bounded, and by Liouville's theorem, it must be constant. If we choose a sequence $z_{\nu }\in G_{0}$ such that $|z_{\nu }|\geq \nu$, the inequality (*) again implies that:
    $\lim \limits _{\nu \to \infty }(z_{\nu })=0,$
thus we conculude that $h\equiv 0$, and in particular $h_{0}\equiv 0$; this is what we wanted to prove

## Conclusions

From the Cauchy integral formula, it follows that every holomorphic function is infinitely differentiable because the integrand in $z$ is infinitely differentiable. We obtain the following results:

### For Circular Disks

Let $G\subseteq \mathbb {C}$ be an open set, $D$ a circular disk with ${\bar {D}}\subseteq G$, and $f\colon G\to \mathbb {C}$ holomorphic. Then $f$ is infinitely differentiable, and for each $n\in \mathbb {N}$, we have
    $f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\partial D}{\frac {f(w)}{(w-z)^{n+1}}},dw$
for each $z\in D$.

### For Cycles

Let $G\subseteq \mathbb {C}$ be an open set, $\Gamma \in C(G)$ a null-homologous cycle, and $f\colon G\to \mathbb {C}$ holomorphic. Then
    $n(\Gamma ,z)\cdot f^{(n)}(z)={\frac {n!}{2\pi i}}\int _{\Gamma }{\frac {f(w)}{(w-z)^{n+1}}},dw$
for each $z\in G\setminus \mathrm {Trace} (\Gamma )$ and $n\in \mathbb {N}$.

## Analyticity

Moreover, every holomorphic function is analytic at every point, i.e., it can be expanded into a power series:

### Statement

Let $G\subseteq \mathbb {C}$ be open, and $f\colon G\to \mathbb {C}$ holomorphic. Let $z_{0}\in G$ and $r>0$ such that ${\bar {B}}_{r}(z_{0})\subseteq G$. Then $f$ can be represented on $B_{r}(z_{0})$ by a convergent power series
    $f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$
where the coefficients are given by
    $a_{n}={\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}},dw$.

### Proof 1

For $z\in B_{r}(z_{0})$, $w\in \partial B_{r}(z_{0})$ we have:
    ${\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}}{(w-z_{0})^{n}}}.\end{array}}$

### Proof 2

the real converges absolutely $|z-z_{0}|<r=|w-z_{0}|$ and we obtain
    ${\begin{array}{rl}f(z)&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{w-z}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\int _{\partial B_{r}(z_{0})}{\frac {1}{w-z_{0}}}{\frac {f(w)(z-z_{0})^{n}}{(w-z_{0})^{n}}}\,dw\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\\&=\displaystyle {\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{\partial B_{r}(z_{0})}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\end{array}}$