Maximum principle for complex valued harmonic function My question is the following:
Is there a maximum principle for functions $u:\Omega\to\mathbb C$, where $\Omega\subset\mathbb{R}^n$ open and bounded and $\Delta u=0$?
More precisely, I'm looking for a statement like
$\Delta u=0\;\Rightarrow\; \max_{\Omega}(|u|) = \max_{\partial\Omega}(|u|)$.
 A: We have such a maximum principle for $\Omega$ bounded and $u$ continuous on $\overline{\Omega}$. It follows from the maximum principle for real-valued harmonic functions:
Since $\overline{\Omega}$ is compact and $\lvert u\rvert$ continuous on $\overline{\Omega}$, $\lvert u\rvert$ attains its maximum at some point $x_0 \in \overline{\Omega}$. If $x_0 \in \partial \Omega$,
$$\sup \:\{ \lvert u(x)\rvert : x \in \Omega\} \leqslant \lvert u(x_0)\rvert = \max\:\{ \lvert u(x)\rvert : x \in \partial \Omega\}$$
follows immediately, and since $u$ is continuous at $x_0$ we also have
$$\lvert u(x_0)\rvert \leqslant \sup\:\{\lvert u(x)\rvert : x \in \Omega\}.$$
If $x_0 \in \Omega$, we conclude that $u$ is constant on the component of $\Omega$ containing $x_0$, and hence $\lvert u\rvert$ also attains its maximum on $\partial \Omega$. If $u(x_0) = 0$, it follows immediately that $u \equiv 0$ on all of $\Omega$. So we can assume $u(x_0) \neq 0$. Let $c\in \mathbb{C}$ with $\lvert c\rvert = 1$ such that $c\cdot u(x_0) > 0$, and define $v = c\cdot u$ and $w = \operatorname{Re} v$. We have $w \leqslant \lvert v\rvert = \lvert u\rvert$, and the real-valued harmonic function $w$ has a local maximum at $x_0$. By the maximum principle for real-valued harmonic functions, $w$ is constant on the component $C$ of $x_0$ in $\Omega$. But then
$$\lvert u(x_0)\rvert = w(x) \leqslant \lvert v(x)\rvert = \lvert u(x)\rvert \leqslant \lvert u(x_0)\rvert$$
on $C$ implies $v = w$ on $C$ and hence $u(x) = u(x_0)$ for all $x\in C$. Since $\varnothing \neq \partial C \subset \partial \Omega$, it follows that $\lvert u\rvert$ also attains its maximum on $\partial\Omega$.
