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I was reading this question. Here we are only considering $\frac{f(z)}{z^n}$ to be bounded for large $|z|$, i.e, $\exists R>0 $ such that $\frac{f(z)}{z^n}$ is bounded for $|z|\geq R$. I was able to understand how we solved this question but I have an additional question.

What will happen if $\frac{f(z)}{z^n}$ is bounded on whole of $\mathbb{C}$?

chesslad
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1 Answers1

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Then $f(z)=kz^n$ for some constant $k$. That's because $0$ is a removable singularity of $\frac{f(z)}{z^n}$ and therefore, near $0$, $\frac{f(z)}{z^n}$ can be written as $a_0+a_1z+a_2z^2+\cdots$. But then, by Liouville's theorem, $a_0+a_1z+a_2z^2+\cdots=k$, for some constant $k$, and therefore $f(z)=kz^n$.

José Carlos Santos
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