Cauchy Integral Formula

Theorem
Let $f \left({z}\right)$ be an analytic function on and within a closed contour $C$.

Let $z_0$ be any point within $C$.

Then:


 * $\displaystyle \oint_C \frac {f \left({z}\right)} {z - z_0} \rd z = 2 \pi i f \left({z_0}\right)$

Proof
Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $f \left({z}\right)$ is analytic.

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


 * $\dfrac {f \left({z}\right)} {z - z_0}$ 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:
 * $\displaystyle I = 2 \pi i f \left({z}\right) + \oint_{C_1} \frac {f \left({z}\right) - f \left({z_0}\right)} {z - z_0} \ \mathrm d z$

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

Hence:

As $\epsilon \to 0$:
 * $\displaystyle \oint_{C_1} \frac {f \left({z}\right) - f \left({z_0}\right)} {z - z_0} \rd x = 0$

So: