Theorem
Let $\Omega\subseteq \mathbb{C}$ be a region and $f\in H(\mathbb{C})$ ($f$ is analytic on $\mathbb{C}$). If $|f|$ has in $\Omega$ a local maximum, then $f$ is constant.
Proof. Let $ D(a,\ r)\subseteq\Omega$ be a disk with $|f(z)|\leq|f(a)|$ for all $z\in D(a,\ r)$ ($a$ is therefore a local maximum for $f$). Then we have for every $\rho\in(0,\ r)$ with $\gamma(t)=a+\rho e^{it} \ (t\ \in[0,2\pi])$:
$$
|f(a)|=\frac{1}{2\pi}\left|\int_{\gamma}\frac{f(\xi)}{\xi-a}\mathrm{d}\xi\right|\ \leq\frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t\ \leq|f(a)| \ \ \ (1)
$$
$\color{green}{\text{I understand the inequality in (1), but what is the point of showing: } |f(a)|\leq|f(a)|?}$
$\color{green}{\text{Why is this mentioned?}}$
From $|f(a+\rho e^{it})|\leq|f(a)$ for $t \in[0,2\pi]$ and $\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$ (2) we have:
$$
|f(a+\rho e^{it})|=|f(a)|\ (t\ \in[0,2\pi]).\ \ \ (3)
$$
$\color{green}{\text{What is the role of the integral in (2), in showing the equality (3)?}}$
Therefore we have:
$$
|f(z)|=|f(a)|\ (z\in D(a,r))\ (*)
$$
If $f$ would not be constant, then (because $\Omega$ is a region) also $f:D(a, r)\rightarrow \mathbb{C}$ would not be constant and $f(D(a,\ r))$ would be open, which is a contradiction to $(*)$.
$
\square
$
===
I still don't get something:
So we know that:
$|f(a+\rho e^{it})|\leq|f(a)$ for $t \in[0,2\pi]$ and
$\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$
I still don't see why:
$f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$ follows.
Best Answer
The point of inequality (1) is that we have equality two. Define $g\colon [0,2\pi]\to\Bbb R$ by $g(t)=|f(a)|-|f(a+\rho e^{it})|$. This function is non-negative, continuous and its integral is $0$ hence $g$ is identically $0$ (that's why we used integral).