Let $f$ be holomorphic on a domain $D$. Suppose $|f(z_0)|$ is a local maximum at some $z_0 \in D$ (interior point). We prove $f$ must be constant in a neighborhood of $z_0$.

Step 1: Mean value property.

By Cauchy's integral formula, for any $r > 0$ small enough that the disk $|z - z_0| \leq r$ lies in $D$:

$f(z_0) = \frac{1}{2\pi i} \oint_{|z-z_0|=r} \frac{f(z)}{z - z_0} dz.$

Parametrize the circle by $z = z_0 + re^{i\theta}$ for $\theta \in [0, 2\pi]$. Then $dz = ire^{i\theta} d\theta$ and

$f(z_0) = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z_0 + re^{i\theta})}{re^{i\theta}} \cdot ire^{i\theta} d\theta = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta.$

This is the mean value property: $f(z_0)$ equals the average of $f$ on the circle $|z - z_0| = r$.

Step 2: Use the hypothesis that $|f(z_0)|$ is maximal.

Taking moduli in the mean value property:

$|f(z_0)| = \left|\frac{1}{2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) d\theta\right| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0 + re^{i\theta})| d\theta.$

By hypothesis, $|f(z_0 + re^{i\theta})| \leq |f(z_0)|$ for all $\theta$. Thus

$|f(z_0)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(z_0)| d\theta = |f(z_0)|.$

Equality holds in the triangle inequality. This forces $f(z_0 + re^{i\theta})$ to have constant argument (all values point in the same direction in the complex plane). Moreover, $|f(z_0 + re^{i\theta})| = |f(z_0)|$ for all $\theta$.

Step 3: Conclude $f$ is constant on the circle.

Write $f(z_0 + re^{i\theta}) = |f(z_0)| e^{i\phi(\theta)}$ where $\phi(\theta)$ is continuous. The mean value property gives

$f(z_0) = \frac{|f(z_0)|}{2\pi} \int_0^{2\pi} e^{i\phi(\theta)} d\theta.$

For this integral to equal $f(z_0) = |f(z_0)| e^{i\phi_0}$ (for some fixed $\phi_0$), we need

$\int_0^{2\pi} e^{i\phi(\theta)} d\theta = 2\pi e^{i\phi_0}.$

This happens if and only if $\phi(\theta) = \phi_0$ for all $\theta$ (the only way a continuous function on $[0, 2\pi]$ can have average value $e^{i\phi_0}$ and modulus $1$ everywhere is to be constant). Thus $f(z_0 + re^{i\theta}) = |f(z_0)| e^{i\phi_0} = f(z_0)$ for all $\theta$.

Step 4: $f$ is constant on the disk.

We have shown $f$ is constant on the circle $|z - z_0| = r$. Since $r > 0$ was arbitrary (any small enough $r$), $f$ is constant on all circles centered at $z_0$ within $D$. By continuity, $f$ is constant in the open disk $|z - z_0| < r_0$ for some $r_0 > 0$.

Step 5: $f$ is constant on $D$ by analytic continuation.

Since $f$ is constant in a neighborhood of $z_0$, all derivatives of $f$ vanish at $z_0$. By the identity theorem for holomorphic functions, $f$ is constant on the entire connected domain $D$.