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This question already has an answer here:

Let $f:\mathbb{C}\to\mathbb{C}$ be an entire function.

Let $n\in\mathbb{N}$ and suppose that

$$\forall w\in\mathbb{C}:\#\{z\in\mathbb{C}:f(z)=w\}\leq n$$

In words, every complex value is attained by $f$ in at most $n$ different places.

Prove that $f$ is a polynomial of degree at most $n$.

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marked as duplicate by Martin R, Namaste, Saad, Will Fisher, Xander Henderson Jun 21 '18 at 2:46

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    $\begingroup$ See math.stackexchange.com/questions/287683 for an approach avoiding Picard's great theorem. $\endgroup$ – mrf Jun 20 '18 at 15:09
  • $\begingroup$ @mrf Your proof at that link is very nice. Don Fanucci, take notice. $\endgroup$ – zhw. Jun 20 '18 at 21:28
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This follows from Picard's great theorem.

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  • $\begingroup$ By considering $\frac 1 f$ ? $\endgroup$ – nicomezi Jun 20 '18 at 12:14
  • $\begingroup$ No. Considering $f\left(\frac1z\right)$. $\endgroup$ – José Carlos Santos Jun 20 '18 at 12:15
  • $\begingroup$ Ho, now I see. Tend to forgot this surprising theorem, may be because I find it hard understanding it. [+1] $\endgroup$ – nicomezi Jun 20 '18 at 12:18
  • $\begingroup$ Oh yes I see, thank you. But, by considering $f(\frac{1}{z})$ I can see that if $f$ is not a polynomial then $z=0$ is an essential singularity, which brings a contradiction to the assumptions using Picard's. But, this only gives that $f$ is a polynomial (without a bound on the degree), then we need to use the local mapping theorem, right? $\endgroup$ – Don Fanucci Jun 20 '18 at 12:22
  • $\begingroup$ You can use the fondamental theorem of algebra for the bound, now that you know that $f$ must be a polynomial. $\endgroup$ – nicomezi Jun 20 '18 at 12:23

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