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This question already has an answer here:

I want to show that a function $f$ entire s.t.

f(z+1)=f(z)=f(z+i)

for all $z$, must be constant.

I think what I need to do is show that the function is bounded in some region of the complex plane, and then by periodicity it must be true that the function is bounded everywhere, and hence by Liouville must be constant. Can someone help me do this? Thanks!

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marked as duplicate by mrf complex-analysis Aug 28 '16 at 9:33

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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    $\begingroup$ That's doubly periodic. The distinction is crucial. $\endgroup$ – zhw. Aug 27 '16 at 22:39
  • $\begingroup$ Note that $f(\mathhbb{C}) = f(\{ a + bi : a, b \in [0, 1] \})$, and that's a compact set, so $f$ is bounded. $\endgroup$ – AJY Aug 27 '16 at 23:22
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Let $z\in C, z=u+vi, u,v\in R$, you can write $u=a+n, v=b+m, 0\leq a,b\leq 1$, $f(z)=f(a+n+(b+m)i)=f(a+ib)$, this implies that $f$ is bounded since its image is contained in the image of the compact space $\{a+ib, a,b\leq 1\}$. We deduce that $f$ is constant.

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Consider the closed disk: $D= \{ z : |z| \leq 2 \}$. Since it is closed and bounded, it must be compact.

Extreme Value Theorem: Continuous functions attain maximum and minimum values on compact domains.

So, $f$ attains its minimum $m$ and maximum $M$ on $D$.

Let $z= a + ib$ for $a,b \in \mathbb R$, and $w = \big(a-\lfloor a \rfloor \big) + i \big(b-\lfloor b \rfloor \big)$.

Observe that $w \in D$ and $f(z) = f(w)$. So, for any $z \in \mathbb C$, we have $m \leq f(z) \leq M$.

Therefore $f$ is bounded.

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