TITLE: If $f$ is an entire function and $f(z) \not \in [0,1]$ for every $z$, then $f$ is constant
QUESTION [11 upvotes]: I want to prove that if $f$ is an entire function and $f(z) \not \in [0,1]$ for every $z$, then $f$ is constant.
If it was written that $|f(z)| \not \in [0,1]$ I would have used the fact that $\frac{1}{f}$ is entire function and Cauchy's formula to bound it.
However it is not the case here, so I don't really see what to do.
I can still use the fact that $\frac{1}{f}$ is also entire, but appart from that, I am clueless.
Help would be appreciated

REPLY [3 votes]: Picard's little theorem: A non-constant entire function $f$ takes all values of complex plane with at most a single exception.
Further, if it misses more than one value then $f$ is constant.

For proof:Picard's Little Theorem Proofs


$ Added$:Define $g:\mathbb C\rightarrow \mathbb C$ as $g=\frac{1}{f}$. Clearly, $g$ misses at least two values ( 1 and 2 say) on $\mathbb C$ (codomain plane),so $g$ must be constant by Picard's little theorem and so is $f$.