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Let $\displaystyle f$ be an entire function such that $$\lim_{|z|\rightarrow \infty} |f(z)| = \infty .$$ Then, Prove that $f$ is polynomial.

My attempts: I was thinking about $f(z) = \sin z$ but it is not polynomial as I am confused.

Please help me thanks in advance.

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  • $\begingroup$ Moreover, $|\sin(x)|\leq 1$ for $x\in\mathbb R$. Hence $|\sin(z)|\to\infty$ for $|z|\to\infty$ doesn't hold. $\endgroup$ – Mundron Schmidt Feb 28 '18 at 8:53
  • $\begingroup$ @Winther: You are right. I falsely interpreted that the converse was true. $\endgroup$ – MrYouMath Feb 28 '18 at 9:33
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Hint: let $g(z)=f(1/z)$ for $z \ne 0$.

If $f(z)= \sum_{n=0}^{\infty}a_nz^n$ for $z \in \mathbb C$, then

Then $g(z)=\sum_{n=0}^{\infty}\frac{a_n}{z^n}$ for $z \ne 0$.

From $\lim_{|z|\rightarrow \infty} |f(z)| = \infty $, we see that $g$ has a pole at $z =0$.

Can you proceed ?

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Hint: for some $r > 0$, $0 < |z| < r\implies f(1/z)\ne 0$ (why?). Let be $h(z) = 1/f(1/z)$ for $0 < |z| < r$, $h(0) = 0$. You can prove easily that $h$ is holomorphic in the disk $|z| < r$ and has a zero of order $m > 1$ at $z = 0$, i.e.,$1/h$ has a pole of order $m$ at $z = 0$. How is the Laurent series of $1/h(z) = f(1/z)$ at $z = 0$?

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