Maximum Modulus Principle

Theorem
A non-constant analytic function $f$ in an open connected set $D$ does not have any interior maximum points.

That is, for each $z \in D$ and $\delta > 0$, there exists some $\omega \in B_\delta \left({z}\right) \cap D$, such that $|f(\omega)| > \left\vert{f \left({z}\right)}\right\vert$.

Proof
Pick some $r > 0$ such that $B_r \left({z}\right) \subset D$.

By the mean value theorem for holomorphic functions, we have:
 * $\displaystyle f \left({z}\right) = \dfrac 1 {2 \pi} \int_0^{2 \pi} f \left({z + r e^{i \theta} }\right) \ \mathrm d \theta$

Then:
 * $\displaystyle \left\vert{f \left({z}\right)}\right\vert \le \frac 1 {2 \pi} \int_0^{2 \pi} \left\vert{f \left({z + r e^{i \theta} }\right)}\right\vert \ \mathrm d \theta \le \max_\theta \left\vert{f \left({z + r e^{i \theta} }\right)}\right\vert$

So it must be that $\exists\omega \in C_r \left({z}\right)$ (circle of radius $r$ centered at $z$) such that $\left\vert{f \left({z}\right)}\right\vert \le \left\vert{f \left({\omega}\right)}\right\vert$.

Note that equality is only obtained when $|f|$ is constant on $C_r \left({z}\right)$.

However, since this holds for all sufficiently small $r > 0$, $|f|$ would be constant in $B_r \left({z}\right)$.

Then $f$ must be constant in $D$, contradicting our assumption.

It follows that $\exists\omega \in C_r \left({z}\right)$ such that:
 * $\left\vert{f \left({z}\right)}\right\vert < \left\vert{f \left({\omega}\right)}\right\vert$