How can we proof the following argument:
Given an analytic function $f(z)$ ($z=x+iy, x,y\in \mathbb R$) in the closed upper half plane such that $|f(z)|<1$ for all $z$ in the open upper half plane, and $|f(x)|=1$ for all $x\in \mathbb R$, then:
there is an open simply connected set $S$ with $\mathbb R \subseteq S \subseteq \mathbb C$ such that $f(z)$ does not vanish on $S$, and there exists an analytic function $\psi:S \rightarrow \mathbb C$ such that $f(z)=\exp(i\psi(z))$