Let $\Omega \subseteq \mathbb R^n$ be an open, connected domain, and let $u: \Omega \to \mathbb C$ be (complex-)harmonic on $\Omega$. Furthermore, let $|u|$ have a (global) maximum in some point $x_0 \in \Omega$, i.e. we have $u(x_0) = \sup_\Omega u$. I now want to show that $u$ is constant on all of $\Omega$.
Now my approach was to somehow apply the (strong) maximum principle which tells me that any subharmonic function $\Omega \to \mathbb R$ on a connected open set $\Omega \subseteq \mathbb R^n$ that has a maximum within $\Omega$ is constant. My problem is that in the above preliminaries, I have a harmonic function $\Omega \to \mathbb C$, so I think I have to apply the maximum principle for its absolute value $|u|: \Omega \to \mathbb R$ instead, but do I know in this given context that $|u|$ is subharmonic, and if so, how could I see/deduce that?
We get the subharmonicity of $\lvert u\rvert$ from the Poisson integral, for example. If $B \subset \Omega$ is a ball, for $x\in B$ we have
$$u(x) = \int_{\partial B} P(\xi,x)u(\xi)\,d\sigma(\xi)$$
where $P$ is the Poisson kernel for the ball $B$ and $\sigma$ the surface measure on the boundary sphere. Since $P \geqslant 0$, taking absolute values yields
$$\lvert u(x)\rvert \leqslant \int_{\partial B} P(\xi,x) \,\lvert u(\xi)\rvert\,d\sigma(\xi)$$
for all $x\in B$. Thus $\lvert u\rvert$ is subharmonic.
Another way to prove the assertion is to multiply $u$ with a complex number $\alpha$ of absolute value $1$ so that $\alpha\cdot u(x_0) \geqslant 0$. Since $v := \operatorname{Re} (\alpha u)$ is a real-valued harmonic function, the maximum principle for real-valued harmonic functions implies that $v$ is constant, and from that we deduce that $u$ is constant (since $\operatorname{Im} (\alpha u(x)) \neq 0$ for some $x$ would contradict the assumption that $\lvert u(x_0)\rvert$ is maximal).