TITLE: If $f(\mathbb{C})\subset \mathbb{C}-[0,1]$ then $f$ is constant
QUESTION [6 upvotes]: If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is an entire function such that $f(z)\neq w$ for all $z\in \mathbb{C}$ and for all $w\in [0,1]\subset \mathbb{R}$, how to prove that $f$ is constant (without using Picard's little theorem). 
Any hint would be appreciated.

REPLY [3 votes]: First show that there is an (injective) holomorphic mapping from
$\Bbb C \setminus [0, 1]$ into the unit disk:

Start with $z \to 1 - \frac 1z$, this maps $\Bbb C \setminus [0, 1]$
into $\Bbb C \setminus (-\infty, 0]$.
Continue with the principal branch of the square root which maps
$\Bbb C \setminus (-\infty, 0]$ to the right half-plane.
Finally, $z \to \frac{z-1}{z+1}$ maps the right half-plane to the
unit disk.

Denoting the composition of these mappings with $\phi$ , $\varphi \circ f$ is a bounded holomorphic function.
Liouville's theorem implies thtat $\varphi \circ f$ is constant.
Now conclude that $f$ is constant as well.