I’ve been reviewing for my Complex Analysis final, and making sure I know how to prove the big theorems we’ve covered. I figured I might like to post this on my website so that I can make sure I have my understanding down pat.

Statement of the Theorem

Let $f$ be an analytic function on an open set $U \subseteq \mathbb{C}$. Let $W \subseteq U$ be relatively compact. Then if $a \in W$ and $r > 0$ such that $D(a, r)$¹ $\subseteq W$, we have $f(a) = \frac{1}{2 \pi i} \int_{\partial W} \frac{f(z)}{z-a} dz$.

Proof

Define $W_r = W \setminus D(a, r)$. Then the function $\frac{f}{z-a}$ is analytic on $W_r$, since $a \not\in W_r$. Now let’s consider the integral $\int_{\partial W_r} \frac{f(z)}{z-a} dz.$ By Stokes’ Theorem, we have that \[\int_{\partial W_r} \frac{f(z)}{z-a} dz = \int_{W_r} d \left(\frac{f(z)}{z-a} dz \right).\] But this is $0$ because $\frac{f(z)}{z-a}$ is analytic. Now, \[0 = \int_{\partial W_r} \frac{f(z)}{z-a} dz = \int_{\partial W} \frac{f(z)}{z-a} dz - \int_{|z-a|=r} \frac{f(z)}{z-a} dz.\]

So we just need to show that $\int_{|z-a|=r} \frac{f(z)}{z-a} dz = 2 \pi i f(a)$. Parametrize $|z-a|=r$ as $z = a + r e^{i \theta}$ for $\theta \in [0, 2 \pi]$. Then we have $dz = i r e^{i \theta} d\theta$. Our integral becomes \[i \int_0^{2 \pi} f(a + r e^{i \theta}) d\theta.\]

Now notice that $f$ is continuous since it’s analytic, and since $W$ was relatively compact, we have that $f(a + r e^{i \theta})$ converges uniformly to $f(a)$ as we take $r \to 0$. Since our integral expression holds for any $r$ small enough, we pass through the limit to find that \[\int_{\partial W} \frac{f(z)}{z-a} dz = i \int_0^{2 \pi} f(a)d\theta = 2 \pi i f(a),\] which proves the theorem. $\square$

Note that while we proved this theorem, we also proved Gauss’ MVT!

Thanks to Mihai Putinar for the proof from MATH CS 122A.