If $f(\mathbb{C})\subset \mathbb{C}-[0,1]$ then $f$ is constant 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.
 A: $f(\Bbb C)\subset \Bbb C\setminus [0,1]$, take $g=1-\dfrac{1}{f}$, $g$ is an entire function and $g(\Bbb C)\subset \Bbb C\setminus \Bbb R_-$, now we can composite with $\log$, and we obtain an entire function $h=\log \circ g$.
$h$ is an  entire function and $h(\Bbb C)\subset \Bbb R\times ]-\pi, \pi [$.
To finish remark that the function $z\mapsto \dfrac{1}{h(z)-3\pi i}$ is an entire function and it is bounded, by the Liouville's theorem it is constant, hence $f$ is constant. 
A: 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.
