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In a recent question that was just deleted, @danielfischer gave at the end of his answer the following exercise: for entire $f$, $$e^{f(z)} \text{ is a polynomial} \iff f \text{ is constant}$$

I was thinking about how to prove this...my first idea was to use that $\text{Re}(f)$ is majored by some $\log$ function as $z\to \infty$. Any ideas?

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@danielfischer ---- –  Eric Auld Dec 21 '13 at 15:51

2 Answers 2

up vote 6 down vote accepted

Hint: Consider the type of the singularity of $e^{f(z)}$ in $\infty$.

Different hint: Does $e^{f(z)}$ have any zeros?

The first hint generalises to show that $e^{h(z)}$ for analytic $h$ never can have a pole in an isolated singularity of $h$.

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Thank you very much. I see the second way, and I'm going to chew on the first way to make sure my understanding of it is rigorous. (I think Taylor expand $\exp(f(1/z))$ at zero.) –  Eric Auld Dec 21 '13 at 16:08
    
I wouldn't mess with Taylor expansions. $f$ itself has an isolated singularity in $\infty$. Use the Casorati-Weierstraß theorem and the mapping properties of $\exp$ to conclude that $e^f$ cannot have a pole. –  Daniel Fischer Dec 21 '13 at 16:11

It can be solved using more elementary tools.

It $\mathrm{e}^{f(z)}$ is a non-constant polynomial, then it has a root (due to the Fundamental Theorem of Algebra), which contradicts the fact that the exponential does not vanish.

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