Pre-image of entire functions

Question

Let $f$ be an entire function which is not a polynomial, and $a \neq b$. Can $f^{-1}(\{a,b\})$ be a finite set?

Answer 1

$f(z)=z$ has this property since the inverse image has at most two points.

If $f$ is not a polynomial then $f$ has an essential singularity at $\infty$. By Picard's theorem $f$ attains each value, except possibly one, at inifinitely many points.

[Is $f(\frac 1 z)$ has a removable singularity or a pole at $0$ then $z^{n} f(\frac 1 z)$ is analytic at $0$ for some $n$. Hence $\frac {f{(z)}} {z^{n}}$ is bounded. This implies that $f$ is a polynomial].

Answer 2

Hint: the Great Picard Theorem.