Solution: Let $z_0 \in \Omega $ be a local maxima of $\vert f \vert$. Since $f$ is analytic in $\Omega $ consider its power series around $z= z_0$, \[ f(z) = \sum_{n=0}^{\infty} a_n \left( z - z_0 \right) ^n,\quad a_n \in \mathbb{C} . \] Then, $f(z_0) = a_0$ and as $r \rightarrow 0$ due to continuity of $f$, for every $\theta \in \mathbb{R} $ \[ f\left( z_0 + r e^{\iota \theta } \right) \rightarrow f\left( z_0 \right) = a_0. \] Therefore, for small enough $r$, and for any $\theta \in \mathbb{R} $ \begin{equation}\label{eq:20Aug2024-1} \left\vert f\left( z_0 + r e^{\iota \theta } \right) \right\vert^2 \leq \left\vert a_0 \right\vert ^2. \end{equation}

Now we recall that Parseval's identity for the above function, \[ \frac{1}{2\pi } \int _0^{2\pi } \left\vert f\left( re^{\iota \theta } \right) ^2 \right\vert \mathrm{d} \theta = \sum_{n=0}^{\infty} \vert a_n \vert ^2 r^{2n},\quad 0 < r < R, \] where $R$ is the radius of convergence of the power series of $f$. Therefore, \begin{align*} \int _0^{2\pi } \left\vert f\left( z_0 + r e^{\iota \theta} \right) \right\vert \mathrm{d} \theta & = 2\pi \sum_{n=0}^{\infty} \vert a_n \vert ^2 r^{2n} . \end{align*} Also, using \eqref{eq:20Aug2024-1}, we have \begin{align*} \int _0^{2\pi } \left\vert f\left( z_0 + r e^{\iota \theta} \right) \right\vert \mathrm{d} \theta \leq 2\pi \vert a_0 \vert ^2. \end{align*} Thus, for sufficiently small $r$ \begin{align*} 2\pi \sum_{n=0}^{\infty} \vert a_n \vert ^2 r^{2n} \leq 2\pi \vert a_0 \vert ^2 , \end{align*} which implies $a_1 = a_2 = \dots = 0$, and hence $f$ is constant, a contradiction.