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Let $f$ be an entire function s.t $f(\mathbb{C}) \cap B_R(z_0) = \varnothing$ for some $z_0$ and some $R$. Then $f$ is constant.

I guess since the image of the whole plane isn't dense then $f$ doesn't have an essential singularity at infinity. Now I have to exclude it has a polar singularity at infinity..

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Please edit the title so that it does not sound like you want to prove a false statement! – Mariano Suárez-Alvarez Apr 17 '12 at 20:32
To give you an idea as to how much 'surjectivity' does an entire, non-constant function have, take a look at the Picard theorem: – Beni Bogosel Apr 17 '12 at 20:53
The other answers of cause work very well, but you could also continue your own. You excluded the case of an essential singularity at infinity. Then it has no pole at all or a polar singularity at infinity. Hence it is a polynomial and thus is either constant or surjective. – Ben Apr 18 '12 at 19:01
Great! Thank you – balestrav Apr 18 '12 at 22:56
up vote 11 down vote accepted

Let $g(z):=\frac 1{f(z)-z_0}$, then $g$ is well-defined and analytic. We have $|g(z)|\leq\frac 1R$ (because $|f(z)-z_0|\geq R$ so $g$ is bounded and hence; by Liouville's theorem, constant. So $f$ is constant.

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Consider this map: $$z\mapsto\frac{1}{f(z)-z_0}$$

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