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Clearly, $f(z)=z^2e^z$ is one such function. Suppose $g$ is another entire function satisfying the given criterion, then $|f/g|=1$ when $z\neq0$. I want to invoke Liouville's theorem and conclude all such functions are given by $f(z)e^{i\theta}$ for any fixed real number $\theta$. Since $f$ and $g$ are entire and nonzero when $z\neq0$, the ratio $f/g$ is analytic in $\mathbb{C}-\{0\}$.

I think I'm on the right track but unsure how to make the above argument precise at $z=0$.

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3  
Have you seen Riemann's theorem on removable singularities? –  Dylan Moreland Aug 16 '11 at 20:00
1  
(Liouville, not Louisville.) –  Hans Lundmark Aug 16 '11 at 20:23
    
I used to be good at spelling, I swear! I don't think I have seen Riemann's theorem before. Thanks! –  dls Aug 16 '11 at 21:23

1 Answer 1

up vote 5 down vote accepted

There is a standard theorem that sends that if $D\subseteq \mathbb{C}$ is open, $z_0\in D$, $f$ is holomorphic on $D\backslash \{ z_0\}$ and bounded on some neighborhood containing $z_0$, then this singularity is removable in the sense that there exist a unique complex number $f(z_0)$ such that the function $$ \widetilde{f}(z)=\begin{cases}f(z_0) & \text{if }z=z_0 \\ f(z) & \text{otherwise}\end{cases} $$ defined on $D$ is holomorphic.

This theorem applies here so that $f/g$ has an entire extension. This extension is bounded, hence a constant, hence $f/g$ is constant.

Keep in mind you are using the fact that the only possible location that $g$ could have a root is $z=0$, so that $f/g$ is automatically analytic everywhere except the origin.

EDIT: Ahh yes, so evidently, this "standard theorem" has a name. See Dylan Moreland's comment :)

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