[proofplan] We show that the set where $|f|$ attains its maximum is both open and closed in $\Omega$, hence all of $\Omega$ by connectedness. Closedness follows from continuity. Openness is the substantial step: at any point $z_0$ where $|f(z_0)|$ is maximal, we apply the Mean Value Property to show that $|f| = |f(z_0)|$ on a surrounding disc. The key tool is the rigidity of the average -- if the modulus of a continuous function achieves its maximum at a point, and the function's value at that point equals its average over a circle, then the modulus must be constantly maximal on that circle. [/proofplan] [step:Handle the case $f(z_0) = 0$] If $f(z_0) = 0$, then $|f(z_0)| \ge |f(z)|$ for all $z \in \Omega$ forces $|f(z)| = 0$ for all $z \in \Omega$, so $f \equiv 0$ on $\Omega$. We may therefore assume $f(z_0) \neq 0$ for the remainder of the proof. [/step] [step:Define the maximum modulus set and establish closedness] Define the maximum modulus set \begin{align*} S := \{ z \in \Omega : |f(z)| = |f(z_0)| \}. \end{align*} By hypothesis $z_0 \in S$, so $S \neq \varnothing$. The function $z \mapsto |f(z)|$ is continuous on $\Omega$ (as $f$ is holomorphic, hence continuous), and $S$ is the preimage of the closed set $\{|f(z_0)|\} \subset \mathbb{R}$ under this continuous map, so $S$ is closed relative to $\Omega$. [/step] [step:Show $S$ is open using the Mean Value Property] We show that $S$ is open in $\Omega$. Let $z_1 \in S$, so $|f(z_1)| = |f(z_0)|$. Since $\Omega$ is open, choose $\rho > 0$ such that $\overline{B}(z_1, \rho) \subset \Omega$. We claim that $B(z_1, \rho) \subset S$. Fix any $r$ with $0 < r \le \rho$. By the [Mean Value Property for Holomorphic Functions](/theorems/3346), since $\overline{B}(z_1, r) \subset \Omega$: \begin{align*} f(z_1) = \frac{1}{2\pi} \int_0^{2\pi} f(z_1 + re^{i\theta})\, d\mathcal{L}^1(\theta). \end{align*} Taking absolute values and applying the triangle inequality: \begin{align*} |f(z_0)| = |f(z_1)| = \left| \frac{1}{2\pi} \int_0^{2\pi} f(z_1 + re^{i\theta})\, d\mathcal{L}^1(\theta) \right| \le \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta). \end{align*} Since $z_1 + re^{i\theta} \in \Omega$ for all $\theta \in [0, 2\pi]$ and $|f(z_0)|$ is the maximum of $|f|$ on $\Omega$, we have $|f(z_1 + re^{i\theta})| \le |f(z_0)|$ for all $\theta$. Therefore \begin{align*} \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta) \le |f(z_0)|. \end{align*} Combining the two inequalities: \begin{align*} \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta) = |f(z_0)|. \end{align*} Define $g: [0, 2\pi] \to \mathbb{R}$ by $g(\theta) = |f(z_0)| - |f(z_1 + re^{i\theta})|$. Then $g \ge 0$ (since $|f| \le |f(z_0)|$ on $\Omega$), $g$ is continuous, and \begin{align*} \int_0^{2\pi} g(\theta)\, d\mathcal{L}^1(\theta) = 2\pi |f(z_0)| - \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta) = 0. \end{align*} A non-negative continuous function with zero integral must vanish identically, so $g(\theta) = 0$ for all $\theta \in [0, 2\pi]$. This gives $|f(z_1 + re^{i\theta})| = |f(z_0)|$ for all $\theta \in [0, 2\pi]$. Since this holds for every $r \in (0, \rho]$, every point of $B(z_1, \rho)$ lies in $S$. Therefore $S$ is open in $\Omega$. [guided] This is the heart of the proof. The strategy is to exploit the rigidity of the mean value integral: if the average of $f$ over a circle has modulus equal to the supremum of $|f|$, then $|f|$ must be constantly equal to that supremum on the circle. Let $z_1 \in S$ and choose $\rho > 0$ with $\overline{B}(z_1, \rho) \subset \Omega$. Fix $r \in (0, \rho]$. Since $\overline{B}(z_1, r) \subset \Omega$ and $f \in \mathcal{O}(\Omega)$, the [Mean Value Property for Holomorphic Functions](/theorems/3346) gives \begin{align*} f(z_1) = \frac{1}{2\pi} \int_0^{2\pi} f(z_1 + re^{i\theta})\, d\mathcal{L}^1(\theta). \end{align*} We now extract modulus information from this equation. Taking absolute values and applying the integral triangle inequality: \begin{align*} |f(z_0)| = |f(z_1)| \le \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta). \end{align*} On the other hand, $|f(z_0)|$ is the maximum of $|f|$ on $\Omega$, so $|f(z_1 + re^{i\theta})| \le |f(z_0)|$ for all $\theta$. Averaging this bound: \begin{align*} \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta) \le |f(z_0)|. \end{align*} The two inequalities force equality: \begin{align*} \frac{1}{2\pi} \int_0^{2\pi} |f(z_1 + re^{i\theta})|\, d\mathcal{L}^1(\theta) = |f(z_0)|. \end{align*} Why does equality force $|f| = |f(z_0)|$ on the circle? Define $g: [0, 2\pi] \to \mathbb{R}$ by $g(\theta) = |f(z_0)| - |f(z_1 + re^{i\theta})|$. We know $g \ge 0$ everywhere (from the maximum assumption) and $\int_0^{2\pi} g\, d\mathcal{L}^1 = 0$ (from the equality just proved). Since $g$ is continuous (as $f$ is continuous and $|\cdot|$ is continuous), a non-negative continuous function integrating to zero must vanish identically. This is a standard fact: if $g$ were positive at some $\theta_0$, by continuity $g > g(\theta_0)/2 > 0$ on an open interval around $\theta_0$, and the integral would be strictly positive, a contradiction. Therefore $|f(z_1 + re^{i\theta})| = |f(z_0)|$ for all $\theta \in [0, 2\pi]$. Since $r \in (0, \rho]$ was arbitrary, every point $z \in B(z_1, \rho)$ satisfies $|f(z)| = |f(z_0)|$, so $B(z_1, \rho) \subset S$ and $S$ is open. [/guided] [/step] [step:Conclude by connectedness that $f$ is constant] The set $S$ is non-empty (since $z_0 \in S$), open in $\Omega$ (by the previous step), and closed relative to $\Omega$ (by continuity). Since $\Omega$ is connected, the only subsets of $\Omega$ that are both open and closed are $\varnothing$ and $\Omega$ itself. Therefore $S = \Omega$, meaning $|f(z)| = |f(z_0)|$ for all $z \in \Omega$. It remains to show that $f$ is constant, not merely that $|f|$ is constant. Since $f(z_0) \neq 0$ (handled in the first step), write $M = |f(z_0)| > 0$. The image of $f$ lies on the circle $\{w \in \mathbb{C} : |w| = M\}$. But $f$ is holomorphic and non-constant holomorphic functions are open maps by the [Open Mapping Theorem for Holomorphic Functions](/theorems/358): the image $f(\Omega)$ would be an open subset of $\mathbb{C}$, which cannot be contained in the circle $\{|w| = M\}$ (a set with empty interior in $\mathbb{C}$). Therefore $f$ must be constant on $\Omega$. [guided] We have shown that $S = \{z \in \Omega : |f(z)| = |f(z_0)|\}$ is both open and closed in $\Omega$, and non-empty. Connectedness of $\Omega$ then forces $S = \Omega$, so $|f|$ is constant on $\Omega$. Does constant modulus imply constant function? Not for arbitrary continuous functions (e.g., $z \mapsto e^{i\theta(z)}$ can have constant modulus without being constant), but for holomorphic functions the answer is yes. We give two ways to see this. **Via the Open Mapping Theorem.** Since $f(z_0) \neq 0$, we have $M := |f(z_0)| > 0$, and $f(\Omega) \subset \{w \in \mathbb{C} : |w| = M\}$. The [Open Mapping Theorem for Holomorphic Functions](/theorems/358) states that if $f$ is a non-constant holomorphic function on a domain, then $f$ maps open sets to open sets. If $f$ were non-constant, $f(\Omega)$ would be an open subset of $\mathbb{C}$. But a subset of the circle $\{|w| = M\}$ has empty interior in $\mathbb{C}$, so it cannot be open. This contradiction forces $f$ to be constant. This completes the proof of the Strong Maximum Modulus Principle: if $|f|$ attains its maximum at an interior point of a domain, then $f$ is constant. [/guided] [/step]