[proofplan] We prove the maximum modulus principle by contradiction. If $|g|$ attains a maximum at an interior point $z_0$, the mean value property of analytic functions forces $|g|$ to be constantly equal to $|g(z_0)|$ on a disk around $z_0$. But a non-constant analytic function with constant modulus on an open set must be constant, yielding a contradiction. [/proofplan] [step:Assume $|g|$ attains a maximum at $z_0 \in \Omega$ and derive constancy on a disk] Suppose $|g|$ attains a maximum at $z_0 \in \Omega$. If $g(z_0) = 0$, then $|g(z)| \leq 0$ on a disk, so $g \equiv 0$ there. Otherwise, let $M = |g(z_0)| > 0$. By the mean value property of analytic functions, for any $0 < \rho < r$ with $D(z_0, r) \subset \Omega$: \begin{align*} M = |g(z_0)| &\leq \frac{1}{2\pi}\int_0^{2\pi} |g(z_0 + \rho e^{i\theta})|\,d\mathcal{L}^1(\theta) \leq M. \end{align*} Equality throughout forces $|g(z_0 + \rho e^{i\theta})| = M$ for all $\theta$ (since a continuous function $\leq M$ whose average equals $M$ must equal $M$ everywhere). This holds for all $0 < \rho < r$, so $|g|$ is constant on $D(z_0, r)$. [/step] [step:Conclude that constancy of $|g|$ on an open set forces $g$ to be constant] A non-constant analytic function with constant modulus on an open disk would satisfy $g\bar{g} = M^2$. By the [open mapping theorem](/theorems/631) (or the Cauchy-Riemann equations), this forces $g$ to be constant, contradicting the hypothesis. Therefore $|g|$ cannot attain a maximum in $\Omega$. [/step]