Proposition

Suppose that $f$ is holomorphic in the region $G$ and $f(G)$ is a subset of the unit circle. Show that $f$ is constant.

Solution

If $f(G) = \{ 1 \}$, we are done. Suppose otherwise. Let $g \in G$ such that $f(g) \ne 1$. Let $U$ be a neighborhood of $g$ in $G$ such that $1 \notin f(U)$. Then let $h: U \rightarrow \mathbb{C}$ such that $h(z) = (1 + f(z)) / (1 - f(z))$. Then $h$ is holomorphic because a composition of two holomorphic functions is holomorphic. As shown in this post, $h$ maps $U$ into the imaginary line. Therefore, $h_x = 0$. By the Cauchy-Riemann equations, $h_y = 0$, so $h$ is constant. Since the Mobius transformation $z \mapsto (1 + z) / (1 - z)$ is not a constant map, $h$ is constant if and only if $f$ is on $U$.

Since $G$ is connected, such $U$’s overlap, so $f$ is constant.