10 - Cauchy's Integral Formula

Cauchy's Integral Formula

Suppose that $f:U\to \mathbb{C}$ is a holomorphic function on an open set $U$
Let $\gamma$ be a simply positively oriented closed curve such that ${\gamma }^{\ast }$ and interior of $\gamma$ are inside $U$

Let $w\in U$

For all $w$ inside of $\gamma$ then

$$f(w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$$

Proof

Fix $w$ inside $\gamma$

Then there exists $r_0$ such that

$$B(w,r_0)\cap {\gamma }^{\ast }=\varnothing$$

So the disc $B(w,r_0)$ is inside $\gamma$

For any $0<r<r_0$ then
Let ${\gamma }_r$ be the positively oriented circle of radius $r$ around $w$

By Homotopy between interior closed paths and interior circle then

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz=\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz$$

As

$$\begin{array}{rl}\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz& =\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)-f(w)}{z-w}dz+\frac{f(w)}{2\pi i}{\int }_{{\gamma }_r}\frac{1}{z-w}dz\\ & =\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)-f(w)}{z-w}+f(w)\end{array}$$

As $f$ is complex differentiable at $z=w$ then

$$\frac{f(z)-f(w)}{z-w}\text{ is bounded as }r\to 0$$

By estimation lemma then the integral over ${\gamma }_r$ tends to $0$

Thus as $r\to 0$ then the path integral around ${\gamma }_r$ tends to $f(w)$

Since $\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-w}dz$ is independent of $r$ then it must be equal to $f(w)$