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 $D$:


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

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 holomorphic on $R$.

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


 * $\dfrac {\map f z} {z - z_0}$ is holomorphic on $R \setminus \set {z_0}$

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

This makes $C$ and $C_r$ a multiply connected region for a sufficiently small $r > 0$.

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

Let:

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

From Complex-Differentiable Function is Continuous, $f$ is continous.

According to Epsilon-Delta definition, for every $\epsilon \in \R_{>0}$ there exists a $\delta \in \openint 0 r$ such that:
 * $\forall z \in \C: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$

Hence:

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

So:

Also see

 * Definition:Complex Contour Integral