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

#### raed

Hi all, I need the solution of the following problem
 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
Hi all, I need the solution of the following problem
 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?