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This question already has an answer here:

Show that if $f$ is a non-constant entire function,it cannot satisfy the condition: $$f(z)=f(z+1)=f(z+i)$$

My line of argument so far is based on Liouville's theorem that states that every bounded entire function must be a constant.

So I try-to no avail-to show that if $f$ satisfies the given condition, it must be bounded. I haven't made much progress with this, so any hints or solutions are welcome.

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marked as duplicate by mrf, Community May 21 '15 at 10:41

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.

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First you can prove that $$f(z+\mathbb Z(i)) = f(z)$$ so $f$ is determined by it's values on the square $[0, 1+i)$. Entire functions are bounded on precompact sets. QED.

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  • $\begingroup$ What is meant here by "entire functions on precompact domains"? By definition, the domain of an entire function is all of $\Bbb C$. $\endgroup$ – Travis May 21 '15 at 8:24
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    $\begingroup$ @Travis Entire functions are holomorphic on every subset of $\mathbb C$. This allows us to conclude boundedness on any precompact subset of $\mathbb C$. $\endgroup$ – AlexR May 21 '15 at 8:26
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    $\begingroup$ This wording is much clearer, cheers! $\endgroup$ – Travis May 21 '15 at 9:24

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