Proving an entire function which misses a ball is constant

complex-analysis

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.


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

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