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Math Help - Entire Functions

  1. #1
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    Entire Functions

    let f be an entire function..

    1) prove if e^f is bounded then f is constant
    2) prove that if Re f is bounded then f is constant

    i'm guessing you would have to use suitable exponentials but i don't have a good enough idea of what to do here. any help would be greatly appreciated xx
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  2. #2
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    Quote Originally Posted by joanne_q View Post
    let f be an entire function..

    1) prove if e^f is bounded then f is constant
    If f is entire then so is e^f, so the result follows from Liouville's theorem.
    Quote Originally Posted by joanne_q View Post
    2) prove that if Re f is bounded then f is constant
    e^f = e^{\text{re}\,f}e^{i\,\text{im}\,f}, so |e^f| = e^{\text{re}\,f}. If re f is bounded then so is e^f, and the result follows from 1).
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    Quote Originally Posted by Opalg View Post
    If f is entire then so is e^f, so the result follows from Liouville's theorem.
    how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

    also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
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    Quote Originally Posted by smoothman View Post
    how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

    also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
    exactly what was on my mind too. any suggestions?
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  5. #5
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    Quote Originally Posted by smoothman View Post
    how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...
    I suppose this just amounts to the chain rule from calculus. If f is differentiable then so is e^f, with \frac d{dz}(e^{f(z)}) = f'(z)\,e^{f(z)}. The definition of an entire function is that it is a function that is everywhere differentiable, and it follows that if f is entire then so is e^f.

    Quote Originally Posted by smoothman View Post
    also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
    Liouville's theorem says that if an entire function is bounded then it is constant. If e^f is constant then so is f. You don't need to bring the Taylor series into it.
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    Quote Originally Posted by joanne_q View Post
    exactly what was on my mind too. any suggestions?
    Quote Originally Posted by smoothman View Post
    how can you say if f is entire then so is e^f... is this just an assumption or is there a proof for this...

    also, if the result follows from louville's theorem, are we meant to show the taylor series for e^f about 0??? where would we go from there.. i would be greatful if you could show me what is to be done here???
    Are you in the same class?
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