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Let $f$ meromorphic function in $\mathbb {C}$. Suppose that there are $R> 0$, $K> 0$ and $n \in {\mathbb {N}}$ such that $\left | {f (z)} \right | \leq {K \left | {z} \right | ^ n}$ for all $\left | {z} \right|> R$. Show that $f$ is a rational function.

DM: Is easily proved that $f$ only has poles at $D =\left\{{z \in {\mathbb {C}}: \left| {z} \right| \leq {R}} \right\}$ belong to a compact bone which implies that $f$ has a finite number of poles, otherwise the set of poles of $f$ has an accumulation point and that is a contradiction. Then $f (z) = \displaystyle \frac{g (z)}{(z-z_1) (z-z_2) ... (z-z_m)}$ where $z_1,... z_m$ are the poles of $f$. My question is if I can narrow the polynomial $p (z) = (z-z_1) (z-z_2) ... (z-z_m)$?. Thank you very much.

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    $\begingroup$ Use dollar signs $\$$ instead of [tex]. $\endgroup$ – user61527 Oct 14 '13 at 4:56
  • $\begingroup$ There is a preview below where you type to see how the question will look while editing. $\endgroup$ – Jonas Meyer Oct 14 '13 at 5:05
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Using your argument, you need only to show that if $f$ is entire and $|f(z)| < K(1+|z|^n)$, then $f$ is a polynomial. That can be shown using Cauchy integral formula. Write $f$ as a power series expansion $\sum a_k z^k$, then you can show $a_k=0$ for all $k> n$.

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