# What can you say about $f(z)$ with finitely many poles?

Assume that $f$ is analytic with finitely many poles {$z_1,z_2,...,z_n$}. At $z=z_i$, $f$ has a pole with multiplicy $m_j>0$. Suppose that $|f(z)\le C(1+|z|)^m$ for $|z|>R$. What can you say about $f$?

If $f$ is entire, then according to the extended Liouville Theorem, we can say $f$ is a polynomial with degree at most $m$. But how do you deal with the poles of $f$?

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Hint: look at the function $$g(z) = (z-z_1)^{m_1} \cdots (z-z_n)^{m_n} f(z)$$
I've thought about this function. But what can you conclude about this function? I know it is entire and bounded in $|z|\le R$. Is it bounded if $|z|>R$? –  user45955 Jan 10 '13 at 20:26
@user45955 No, but it's bounded by $C(1+|z|)^{m+m_1+\cdots+m_n}$. –  mrf Jan 10 '13 at 20:47