complex analysis - using Liouville’s theorem in a proof.

Sep 2009
173
0
Hi all, I need the solution of the following problem
[1] Use Liouville’s theorem to prove that a non-constant polynomial cannot satisfy f(z+1) = f(z) and f(z+i) = f(z) for all z.
Thank for all
 
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Drexel28

MHF Hall of Honor
Nov 2009
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1,566
Berkeley, California
Hi all, I need the solution of the following problem
[1] Use Liouville’s theorem to prove that a non-constant polynomial cannot satisfy f(z+1) = f(z) and f(z+i) = f(z) for all z.
Thank for all
How's this?

Note by induction \(\displaystyle f(z+n)=f(z)=f(z+in)\) and combining these two \(\displaystyle f(z+m+in)=f(z+in)=f(z)\). So, given \(\displaystyle z=x+iy\in\mathbb{C}\) we may note that \(\displaystyle x=\{x\}+\left\lfloor x\right\rfloor, y=\{y\}+\left\lfloor y\right\rfloor\) (fractional and integer part). So, letting \(\displaystyle \lfloor x\rfloor =m,\lfloor y\rfloor =n\) we see that \(\displaystyle f(z)=f(x+iy)=f(\{x\}+m+i(\{y\}+n))=f(\{x\}+\{y\}i)\) an thus \(\displaystyle f\) is entirely determined on \(\displaystyle \Gamma=\left\{x+iy:x,y\in[0,1]\right\}\) but this region is clearly compact and thus \(\displaystyle f\) being continuous is bounded on it. Thus, by previous remarks it follows that \(\displaystyle f\) is bounded on \(\displaystyle \mathbb{C}\) and so \(\displaystyle f(z)=c\)

I don't see why this works for any entire function \(\displaystyle f\) though?