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This question already has an answer here:

I need to apply Liouville theorem ("entire bounded complex functions are constant") to prove that an entire function satisfying:

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

for all complex numbers $z$ is constant. I'm really not sure on how to proceed, I've tried expanding in Taylor series but I've got confused with calculations and I can't proceed any further.

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marked as duplicate by Martin R, user10354138, Paul Frost, Community Dec 9 '18 at 18:06

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By induction, $f(z)=f(z+m+in)$ for all integers $m,n$. This means that for each $z\in\mathbb{C}$ there is $w\in[0,1]\times[0,1]$ such that $f(z)=f(w)$. Continuous functions are bounded on compact sets. It follows that $f$ is bounded on $\mathbb{C}$. Now apply Liouville.

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