Strict Maximum Principle - Complex Version (Gamelin; p88) Let $h$ be a bounded complex-valued harmonic function on a domain $D$. If $|h(z)| \le M$ for all $z \in D$, and $|h(z_0)| = M$ for some $z_0 \in D$, then $h(z)$ is constant on $D$.
First I will copy the proof (then explain where I am confused)
Proof: We replace $h(z)$ by $\lambda h(z)$ for an appropriate unimodular constant $\lambda$, and we can assume $h(z_0) = M$. Let $u(z) = \Re h(z)$. Then $u(z)$ is a harmonic function on $D$ that attains its maximum at $z_0$. By the strict maximum principle for real-valued harmonic functions, $u(z) = M$ for all $z \in D$. Since $|h(z)| \le M$ and $\Re h(z) = M$ we must have $\Im h(z) = 0$ for all $z \in D$ hence $h(z)$ is constant.
Questions
If I understand correctly, we ``redefine'' $h$ as $\lambda h$, so essentially we have $h^* = \lambda h$ for an appropriate $\lambda \in \mathbb{C}$. If that's the case, then shouldn't $u(z) = \Re h^*(z)$, and consequently the maximum scaled by $|\lambda|$??
What if the range of $h$ doesn't include the real-axis, so that we cannot scale $h(z_0)$ to a real value, is that the case?
Thanks in advance
Here $\lambda=|h(z_0)|/h(z_0)$, since $h(z_0)\neq0$ this is well-defined. Moreover $|\lambda|=1$
Set $h^*=\lambda h$ so $h^*(z_0)=|h(z_0)|=M$. Although it is possible that $\operatorname{ran}h$ doesn't include $\mathbb R$, $h^*$ always does.