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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!2012-04-17
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    To give you an idea as to how much 'surjectivity' does an entire, non-constant function have, take a look at the Picard theorem: http://en.wikipedia.org/wiki/Picard_theorem2012-04-17
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    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.2012-04-18

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