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math.stackexchange.com/questions/996450
Since $f$ has a pole at $\infty$ there exists an $R > 0$ such that $|f(z)| \geq 1$ for all $|z| > R$. Now let $z_1, ..., z_k$ be the zeros of $f$, counted with multiplicity, in $\overline{D(0,R)}$. There are only finitely many inside $\{ z: |z| \leq R \}$ or else the zeros would have an accumulation point in this compact set and then $f$ would be identically zero by the uniqueness theorem which is not possible since $f$ has a pole at $\infty$.
Consider the function $h(z) = \frac{(z-z_1)...(z-z_k)}{f(z)}$ then $h$ is an entire function. On the set $\{z: |z| > R \}$, $|h(z)| \leq C R^k$ by the Cauchy estimates since $h$ is an entire function of polynomial growth and hence it must be a polynomial.
On the compact set $\overline{D(0,R)}$, we have $h(z) \leq M$ for some $M>0$ because $h$ is a continuous function on the compact set an hence it attains its maximum there. So $h$ is an entire function that is bounded since $|h(z)| \leq \max \{CR^k, M\}$ on all of $\mathbb{C}$, hence $h$ is a constant function.
But then, $f(z) = h(z)(z-z_1)..(z-z_k)=C(z-z_1)...(z-z_k)$ a polynomial. |
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