# entire function with bounded multiplicity is a polynomial [duplicate]

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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 $n$.

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

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

## 1 Answer

This follows from Picard's great theorem.

• By considering $\frac 1 f$ ? – nicomezi Jun 20 '18 at 12:14
• No. Considering $f\left(\frac1z\right)$. – José Carlos Santos Jun 20 '18 at 12:15
• Ho, now I see. Tend to forgot this surprising theorem, may be because I find it hard understanding it. [+1] – nicomezi Jun 20 '18 at 12:18
• 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? – Don Fanucci Jun 20 '18 at 12:22
• You can use the fondamental theorem of algebra for the bound, now that you know that $f$ must be a polynomial. – nicomezi Jun 20 '18 at 12:23