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I am trying to prove the following:

If $f(z)$ is an entire function on $\mathbb{C}$ and $\displaystyle\lim_{|z|\rightarrow \infty} |f(z)|=\infty$ then $f(z)$ is a polynomial.

Let $a\in \mathbb{C}\setminus \{0\}$. Since $f(z)$ is analytic at $a$, it can be expressed as the power series $$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z-a)^n.$$

I have not learned anything about poles in my complex analysis class, so I am expected to solve this without any theorems involving poles. I am trying to work with the function $g(z):=f(1/z)$. This function would have a singularity at zero, but I won't see how to use the function to prove the claim. Is there a way I can show that there exists some $N$ such that $\frac{f^{(n)}(a)}{n!}=0$ for all $n\geq N$?

Also, prior to this I showed that if $f(z)$ is entire and $|f(z)|\leq A+B|z|^n$ for all $z \in \mathbb{C}$ then $f(z)$ is a polynomial. Perhaps I can invoke this result? Any help is greatly appreciated, thank you.

Ana
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  • Did you learn that at an essential singularity a function doesn't have a limit? Or Picard theorem? – N. S. Apr 28 '17 at 18:43
  • @N.S. Highly doubtful someone who doesn't know about poles knows about the Picard theorems. – Ted Shifrin Apr 28 '17 at 18:48
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    @PedroTamaroff There is no question on this site that solves this problem without invoking a theorem about poles. In fact, I have not learned the Casorati-Weierstrass theorem. – Ana Apr 28 '17 at 18:53
  • @Ana The linked question provides a solution using simple considerations about zeros. In your case, there exists a polynomial $p(z)$ such that $f(z)/g(z)$ is an entire function that is nonvanishing everywhere. In this case, you can write it as $\exp h(z)$ for $h(z)$ entire. Try to use this, which is elementary. – Pedro Apr 28 '17 at 19:00
  • @PedroTamaroff How do you know that such a polynomial exists?? – Ana Apr 28 '17 at 19:18

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