# entire function with only finitely many zeros

I saw the following exercise:

If $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire, non-constant function with only finitely many zeros, then either $|f(z)|\rightarrow \infty$ for $|z|\rightarrow\infty$ or there is a sequence of points $z_n$ such that $|z_n|\rightarrow\infty$ and $f(z_n)\rightarrow 0$.

I thought a bit about this exercise and of course $f$ has to be unbounded because of Liouville's Theorem. But if I assume, that there is a unbounded sequence $z_n$ for which $f(z_n)\rightarrow \infty$ does not hold, how can I conclude, that there has to be a sequence such that $f(z_n)$ goes to zero?

Thanks for hints!

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factor out the zeros and look at the reciprocal – mike May 1 '12 at 16:39

To add on to what Makuasi stated an entire function has a pole of order $n$ at $\infty$ iff the function is a polynomial of order $n$. So if $f$ does not have a pole at $\infty$ $f$ is either bounded or has an essential singularity at $\infty$. $f$ cannot be bounded by Liouville's Thm. So $f$ must have an essential singularity at $\infty$. By Casorati–Weierstrass theorem there is a sequence, $(z_{n})$, such that $|z_{n}| \rightarrow \infty$, and $f(z_{n})\rightarrow 0$
Hint: An entire function $f(z)$ has a pole of order $n$ at $\infty$ iff $f(z)$ is a polynomial of degree $n$ and $f(z)$ is a trancendental entire function then there exist a sequence $z_n$ such that $|z_n|\rightarrow\infty$ for which $f(z_n)\rightarrow\infty$, Let $f(z)=\sum_{k=0}^{\infty}a_kz^k$ be an entire function and have a pole of order $n$ at $\infty$. if we define $g(z)=f(1/z)$, then $g(z)$ has a pole of order $n$ at the origin , It follows that $z^nf(z)$ is bounded near $0$ i.e $z^{-n}f(z)$ is bounded near $\infty$. That is $f(z)$ is entire such that $f(z)\le M|z|^n$ for $|z|>R$. Can you prove from here that $f(z)$ is a polynomial of degree $n$? the converse part is very trivial.