Maximum Modulus Principle

Theorem
A non-constant analytic function $f$ in an open connected set $D$ does not have any interior maximum points: For each $z \in D$ and $\delta > 0$, there exists some $\omega \in B_\delta(z) \cap D$, such that $|f(\omega)| > |f(z)|$.

Proof
Pick some $r > 0$ such that $B_r(z) \subset D$. By the mean value theorem for holomorphic functions, we have

\[ f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z+re^{i\theta})\ \mathrm{d}\theta. \]

Then

\[ \]
 * f(z)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z+re^{i\theta})|\ \mathrm{d}\theta \leq \max_\theta |f(z+re^{i\theta})|.

So it must be that $\exists\omega \in C_r(z)$ (circle of radius $r$ centered at $z$) such that $|f(z)| \leq |f(\omega)|$. Note that equality is only obtained when $|f|$ is constant on $C_r(z)$, but since this holds for all sufficiently small $r > 0$, $|f|$ would be constant in $B_r(z)$. Then $f$ must be constant in $D$, contradicting our assumption. It follows that $\exists\omega \in C_r(z)$ such that $|f(z)| < |f(\omega)|$.