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I would like to see how the following question can be proved:

Let $f\in\mathcal{M}(\mathbb{C})$ satisfying $\vert f(z)\vert\leq M\vert z\vert^n$ for all $z\in\mathbb{C}\setminus P(f)$ with $\vert z\vert>r$ for some finite constants $M,r$ and some $n\in\mathbb{N}$. show that $f$ is a rational function

I tried with Cauchy integral formula and using that $ord(f)=0$ for all $z\in\mathbb{C}\setminus P(f)$ but it didnt get me anywhere

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  • $\begingroup$ You wrote “ for some finite constants $M$, $r$ and some $n\in\mathbb N$”, but where is that $M$? $\endgroup$ – José Carlos Santos Apr 23 at 15:06
  • $\begingroup$ changed it, $\vert f(z)\vert\leq M\vert z\vert^n$ $\endgroup$ – Roni Ben Dom Apr 23 at 15:09
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As $\overline D(0, r)$ is compact, $f$ has a finite number of poles in this disc. So by multiplying by the polynomial $P=\prod (z-\alpha_i)^{m_i}$ for $\alpha_i$ the poles and $m_i$ their multiplicity, you still have $|Pf(z)|\leq |Q|$ for a certain polynomial $Q$, and $z$ outside the disc. But $|Pf|$ is continuous function on this compact disc so we have $C$ such that $|Pf|\leq C$ on the disc. Then, $|Pf|\leq C + |Q|$ everywhere. Then, you can see that there is a polynomial $R$ such that $|Pf|\leq|R|$ everywhere. This is a classical result that an entire function bounded by a polynomial is a polynomial (see for example this thread), so we are done.

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  • $\begingroup$ Why is $\vert Pf\vert$ continuous on the disc? $\endgroup$ – Roni Ben Dom Apr 23 at 16:26
  • $\begingroup$ $|Pf|$ is continuous at each point where $f$ is, so everywhere except at the poles in the disc. At the poles, continuity is assured locally because we are deleting the pole thanks to $P$, by the definition of the multiplicity of a pole $\endgroup$ – elidiot Apr 23 at 16:28

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