Cauchy's Integral Formula

Theorem
Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.

Let $f: U \to \C$ be holomorphic on some open set containing $D$.

Then for each $a$ in the interior of $D$:


 * $\ds \map f a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a} } \rd z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.

Proof
Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is analytic.

Let $z_0$ be any point in the region $R$ such that:


 * $\dfrac {\map f z} {z - z_0}$ is analytic everywhere except at $z_0$.

We draw a circle $C_1$ with center at $z_0$ and radius $r$ such that $r \to 0$.

This makes $C$ and $C_1$ a multiply connected region.

According to Cauchy's Integral Theorem for a multiply connected region:

Let:

Now:
 * $\ds I = 2 \pi i \map f {z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$

According to Epsilon-Delta definition of limit, for every $\left|{z - z_0}\right| < \delta$ there exists a $\epsilon \in \R_{>0}$ such that:
 * $\cmod {\map f z - \map f {z_0} } < \epsilon$

Hence:

As $\epsilon \to 0$:
 * $\ds \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$

So: