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$f$ is an entire function with $|f(z)|\in\mathbb{Z}$ for all $z\in\mathbb{C}$. Is $f$ a constant?

I believe the answer is 'Yes'. Here is my attempt :

Since the assignment $z\to|z|$ is continuous, and $f$ is entire, therefore the composition $z\to f(z)\to|f(z)|$ is continuous. It takes integer values for all $z\in\mathbb{C}$. Hence it has to be constant. Thus, $f$ must be a constant, by Liouville's theorem.

Is this good?

Thank you.

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Hint : The image of a connected set under a continuous function is connected. And the connected subsets of $\mathbb{Z}$ are singletons.

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  • 1
    $\begingroup$ Yeah yeah, I've used this fact. Thanks. $\endgroup$ – WhySee Dec 28 '17 at 13:06
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You have already noted that $g:\mathbb{C}\rightarrow \mathbb{R}$ defined by $g(z)=|f(z)|$ is a continuous function and $g(\mathbb{C})\subseteq\mathbb{Z}$. Since $\mathbb{C}$ is connected so is $g(\mathbb{C})$ and therefore it must be a single integer.

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  • $\begingroup$ Yeah yeah, I've used this fact. Thanks. $\endgroup$ – WhySee Dec 28 '17 at 13:06
  • $\begingroup$ So using Liouville's theorem is not necessary in this context. $\endgroup$ – PJK Dec 28 '17 at 13:10
  • $\begingroup$ Why? For complex numbers $z_1$ and $z_2$, $|z_1|=|z_2|$ does not imply that $z_1=z_2$. Then, to prove that $f$ is a constant, we do need Liouville's theorem, right? $\endgroup$ – WhySee Apr 10 at 12:49
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You have to add the Liouville theorem since $|f(z)|$ is constant implies that $f$ is bounded and a bounded entire function is constant (Liouville).

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  • $\begingroup$ Yeah. I sort of assumed it while writing. I'll make the change. :-) $\endgroup$ – WhySee Dec 28 '17 at 13:00

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