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
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.