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

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    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
    Last edited by mr fantastic; July 3rd 2010 at 05:11 AM. Reason: Re-titled.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by raed View Post
    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 f(z+n)=f(z)=f(z+in) and combining these two f(z+m+in)=f(z+in)=f(z). So, given z=x+iy\in\mathbb{C} we may note that x=\{x\}+\left\lfloor x\right\rfloor, y=\{y\}+\left\lfloor y\right\rfloor (fractional and integer part). So, letting \lfloor x\rfloor =m,\lfloor y\rfloor =n we see that f(z)=f(x+iy)=f(\{x\}+m+i(\{y\}+n))=f(\{x\}+\{y\}i) an thus f is entirely determined on \Gamma=\left\{x+iy:x,y\in[0,1]\right\} but this region is clearly compact and thus f being continuous is bounded on it. Thus, by previous remarks it follows that f is bounded on \mathbb{C} and so f(z)=c

    I don't see why this works for any entire function f though?
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