If $f(\mathbb{C})\subset \mathbb{C}-[0,1]$ then $f$ is constant

$f(\Bbb C)\subset \Bbb C\setminus [0,1]$, 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.

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.