# Entire function $f$ with infinitely many zeros [closed]

Let $f$ be an entire function, non constant with the property: The set $\{w : f(w)=0\}$ has ininite elements. Show that for all c there exist a sequence $\{z_n\}$ such that $z_n\to\infty$ and $f(z_n)\to c$.

Anyone could please give me some help for this exercise? Thanks in advance.

## closed as off-topic by Hanul Jeon, Martin R, Namaste, Jose Arnaldo Bebita-Dris, B. MehtaMay 19 '18 at 16:12

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## 1 Answer

Apply the Casoratti-Weierstrass theorem to $g(z)=f\left(\frac1z\right)$. Note that if the Taylor series of $f$ at $0$ is$$a_0+a_1z+a_2z^2+\cdots,$$then infinitely many $a_n$'s are non-zero, because otherwise $f$ would be either the null function (which is constant) or a non-null polynomial (which has only finitely many zeros). But then the Taylor series of $g$ at $0$ is$$a_0+a_1z^{-1}+a_2z^{-2}+\cdots$$ and therefore $g$ has an essential singularity at $0$.

• Can I assume that g has an essential singularity in 0? – Arturo May 20 '18 at 8:38
• @Arturo No You will have to prove it. But it is not hard. – José Carlos Santos May 20 '18 at 8:47
• So, if I delete the hypothesis "The set {w:f(w)=0} has infinite elements" it will still be truth? I didn't need to use it with the Casoratti-Weirstrass Theorem. – Arturo May 20 '18 at 14:40
• @Arturo But you will need it to prove that $g$ has an essential singularity at $0$. And the statement is false without it. Just tame the identity function. – José Carlos Santos May 20 '18 at 14:50
• @Arturo Do you need a full answer? – José Carlos Santos May 20 '18 at 15:32