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

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

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

2. Originally Posted by raed
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?