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Let $f=u+iv$ be an entire function. Suppose for some $K>0$, we have that $ u > 0$ for all $z$ belonging to $\Omega = \{ z : |z| \geq K \}$. Prove that $f(z)$ is constant.

I guess that I have to use the fact that the image of a non-constant entire function is dense in $\mathbb{C}$, and If I show that $\overline{f(\mathbb{C})} \neq \mathbb{C}$ I can conclude that $f$ is a constant function.... but I do not see how?

The above approach is my main question, but I am also very interested to see other approaches that might work.

Thank you in advance !

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Let $$m:=\inf_{|z| \leq K} u(z) > - \infty. $$ We have that $f$ never takes values in the half plane given by $$\Re w< m .$$ This contradicts the density of the image $f(\mathbb C).$

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  • $\begingroup$ Here like Ayman, you are using the fact that continuous image of a compact set is compact , and hence bounded, so you can define $m$ like that. Am I right ? $\endgroup$ – the8thone Dec 24 '13 at 16:18
  • $\begingroup$ @Roozbeh-unity Correct. $\endgroup$ – user1337 Dec 24 '13 at 16:25
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Since $u$ is continuous, it's bounded on the compact set $\{z \in \Bbb C : |z| \le K\}$. Now consider the function $g(z) = \exp(-f(z))$. Is it bounded? What do you conclude?

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  • $\begingroup$ If I figure out your first sentence, I'm done ! Are you claiming that continuous functions (of 2 variables in particular) whose domain is compact, have compact (I know from topology) and bounded image ? $\endgroup$ – the8thone Dec 24 '13 at 16:13
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    $\begingroup$ @Roozbeh-unity Yes. The image of a compact domain is compact in $\Bbb C$, hence bounded by Heine-Borel. $\endgroup$ – Ayman Hourieh Dec 24 '13 at 16:14

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