Introduction

Let $f:\mathbb{C} \to \mathbb{C}$ be a holomorphic function. By Liouville’s theorem, if $f(\mathbb{C})$ is bounded, then $f$ has to be a constant function. However, there is a much stronger result. In fact, if $f(\mathbb{C})$ differs $\mathbb{C}$ from exactly $2$ points, then $f$ is a constant. In other words, suppose $f$ is non-constant, then the equation $f(z)=a$ for all $a \in \mathbb{C}$ except at most one $a$. To think about this, if $f$ is a non-constant polynomial, then $f(z)=a$ always has a solution (the fundamental theorem of algebra). If, for example, $f(z)=\exp(z)$, then $f(z)=a$ has no solution only if $a=0$.

The proof will not be easy. It will not be proved within few lines of obvious observations, either in elementary approaches or advanced approaches. In this post we will follow the later by studying the twice-punctured plane $\mathbb{C} \setminus\{0,1\}$. To be specific, without loss of generality, we can assume that $0$ and $1$ are not in the range of $f$. Then $f(\mathbb{C}) \subset \mathbb{C}\setminus\{0,1\}$...