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I'm clueless how to start with the solution to this problem.

Fella
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1 Answers1

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First show that $f(z)$ has finitely many zeros (hint: zeros cannot accumulate); then show that this implies that $f(z)$ is a polynomial (both are good exercises about entire functions). The answer then follows easily.

Volanem
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  • To the O.P.: If $f$ is entire and $A\in \mathbb C$ and if ${z: f(z)=A}$ contains an infinite bounded subset then $f(z)=A$ for all $z$. But if $ {z:f(z)=A}$ is unbounded then $|f(z)|$ does not $\to \infty$ as $|z|\to \infty.$ – DanielWainfleet May 23 '17 at 08:45