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Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant.

I want to prove $f$ is bounded. Then by using Liouville theorem, $f$ is constant. But I can't proceed. Can anyone give some hint?

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    $\begingroup$ What can you say about $\lim\limits_{z\to\infty} f(z)$? $\endgroup$ – Daniel Fischer Nov 17 '13 at 14:41
  • $\begingroup$ @Daniel Fischer:The limit exists due to $f(z)=f(\frac{1}{z})$? $\endgroup$ – Idonknow Nov 17 '13 at 18:24
  • $\begingroup$ Right. So $\infty$ is a removable singularity. Hence $f$ is bounded, whence constant. $\endgroup$ – Daniel Fischer Nov 17 '13 at 18:29
  • $\begingroup$ So if $f$ has a removable singularity at $z_0$ and $f$ is entire, then $f$ is bounded? $\endgroup$ – Idonknow Nov 17 '13 at 19:22
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    $\begingroup$ Not any $z_0$. But if an entire function has a removable singularity in $\infty$, that means it can be extended to a holomorphic function on the entire Riemann sphere. The sphere is compact, hence the image id compact, hence bounded. Another way to look at it, if $\infty$ is a removable singularity, then $f$ is bounded on $\{z : \lvert z\rvert > R\}$ for some $R$. The complement of that is compact, hence $f$ is bounded there too, hence $f$ is bounded. $\endgroup$ – Daniel Fischer Nov 17 '13 at 19:27
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Note : If $f$ is entire, then, in particular it is bounded on closed balls around the origin. Hence, if $D = \{z : |z| \leq 1\}$, then $f$ is bounded on $D$ by some $M > 0$.

Hence, if $|z| \geq 1$, then $1/z \in D$, and hence $$ |f(z)| = |f(1/z)| \leq M $$ And hence, $|f(z)| \leq M$ for all $z\in \mathbb{C}$. Hence, Louiville applies.

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    $\begingroup$ The ratio of hences to sentences in your answer exceeds 1. $\endgroup$ – JSchlather Nov 17 '13 at 14:52
  • $\begingroup$ @JSchlather: Mind to elaborate? $\endgroup$ – Idonknow Nov 17 '13 at 16:53
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    $\begingroup$ @JSchlather : That is funny. Hence, I will leave the post as-is. $\endgroup$ – Prahlad Vaidyanathan Nov 17 '13 at 16:58

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