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This question already has an answer here:

I got stuck on this problem. So I really appreciate if anyone can give me some hint to move on. Thanks a lot.

Prove that an entire analytic function $f:\mathbb{C} \rightarrow \mathbb{C}$ is a constant function if $f(z) = f(z+1)=f(z+i)$ for all $z \in \mathbb{C}$

As far as I know, there're some famous theorems on how to prove an entire analytic function is constant, like Liouville theorem, maximal modulus theorem, identity theorem... But I still can't figure out how to apply those to this problem. I think about the dense property of $\mathbb{Q}$ in $\mathbb{R}$, so I tried to prove that $f(z) = f(z+q)$ for all $z \in \mathbb{C}$ and all $q \in \mathbb{Q}$, then use the continuity to prove that it's true for $q \in \mathbb{R}$. But I can't prove that statement.

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marked as duplicate by Martin R, user296602, user147263, colormegone, Shahab Apr 26 '16 at 2:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ and for proving the Liouville theorem, use the Cauchy integral formula : $f^{(n)}(0) = \frac{n!}{2 i \pi} \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz$ which $\to 0$ as $R \to \infty$ (since $|f(z)| < C$) when $n \ge 1$ $\endgroup$ – reuns Apr 25 '16 at 18:29
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The conditions tell that the function is $(1,1)$-periodic and hence $$\sup_{\mathbb{C}}|f| = \max_{x,y\in [0,1]}|f(x + iy)| < \infty.$$

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    $\begingroup$ For the last statement, we can conclude that from the fact that $[0,1] \times [0,1]$ is compact, so ${f([0,1] \times [0,1])}$ is compact, too. So it's bounded, is that right? Then we use Liouville theorem to get the conclusion. Am I wrong at any point? $\endgroup$ – le duc quang Apr 25 '16 at 17:53
  • $\begingroup$ @leducquang: that's correct! $\endgroup$ – Giovanni Apr 25 '16 at 17:55
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    $\begingroup$ So great and simple. Thanks a lot, @Giovanni :-D $\endgroup$ – le duc quang Apr 25 '16 at 17:56
  • $\begingroup$ @leducquang: you are welcome! btw, zhw's answer is pretty much the same: he was pointing out the periodicity in my opinion, which then leads to the consideration in my answer :) $\endgroup$ – Giovanni Apr 29 '16 at 3:52
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Hint: For all $m,n\in \mathbb Z,$ $f(z) = f(z+m+ni).$

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    $\begingroup$ Can you make your point a little clearer? Is it the same solution with @Giovanni? $\endgroup$ – le duc quang Apr 25 '16 at 17:54

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