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Math Help - proof a complex function is zero

  1. #1
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    proof a complex function is zero

    f is a holomorphic function at every z in C. We know that lim f(z)=0 when z->+00. Show that f(z)=0 for every Z in C.
    Can you please give me some kind of hint ?? i would be sooo grateful...
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  2. #2
    GJA
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    Re: proof a complex function is zero

    Hi Eleni,

    Since f(z) is holomorphic at each point of \mathbb{C}, f is an entire function. The idea here is to use Liouville's Theorem. Try finding a way to use the information that f(z)\rightarrow 0 to prove that f is bounded. Once you have f bounded and entire Liouville's Theorem will tell you that f is constant.

    Does this get things on the right track? Good luck!
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  3. #3
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    Re: proof a complex function is zero

    GJA - I don't see a way to easily prove that f is bounded. You can't just use the fact that f(z) goes to zero - there are functions that go to zero but are not bounded.

    I think you can expand f(z) in a power series and use Cauchy's Integral Formula to prove all the coefficients are zero except for the constant coefficient. So f(z) is a constant which must be zero. That's the way the proof of Liouville's Theorem goes.

    I think it might be possible to argue directly from Cauchy's Integral Formula. If so, that would be a much better solution.

    - Hollywood
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  4. #4
    GJA
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    Re: proof a complex function is zero

    If f(z)\rightarrow 0 as |z|\rightarrow\infty, then there is a compact disc centered at the origin such that |f(z)|<1 for z outside of this disc. Since this disc is compact and f is entire (so continuous), f is bounded on the disc. The bound for f is then the maximum of 1 and the bound on the disc.
    Thanks from hollywood
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  5. #5
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    Re: proof a complex function is zero

    Sure enough. Thanks.

    Eleni: GJA's post shows that f is bounded, and you know f is entire. So by Liouville's Theorem, f is constant. Since its limit is zero, then, it must be identically zero.

    - Hollywood
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