Show that a non-constant entire function has a dense image.

Let $f$ be a nonconstant entire function and $U$ be an open set in the plane. Show that there is a $z_0$ such that $f\left(z_0\right)\in U$.

This question is an exercise for the Maximum Modulus and Mean Value section. I can't figure out how to prove this. I'm more than sure this requires an application of the mean value theorem, but I don't exactly know how to use it.

Any suggestions/tips on how to proceed?

• Note that this does not imply that $f$ is actually surjective, only that it has dense image. $\exp$ for instance is not surjective, but its image is the punctured plane, which is dense. – Ian Dec 9 '14 at 0:21
• Oh wow I didn't see that! I see now why I originally thought that and why I was wrong. – Arturo don Juan Dec 9 '14 at 0:33

Hint: consider $1/(f(z) - u)$, and use Liouville's Theorem.
• Could you explain what $u$ is? Thanks for the response by the way. – Arturo don Juan Dec 9 '14 at 0:16
• Suppose $f(z) \notin U$ for all $z$. Pick $u \in U$, then $f(z)$ is bounded away from $u$... – copper.hat Dec 9 '14 at 0:20