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Math Help - another hard complex analysis question

  1. #1
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    another hard complex analysis question

    Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

    Edgar

    Let f be an entire function.
    (a) Prove that if e^f is bounded then f is constant.
    (b) Prove that if Re f is bounded then f is constant.
    Hint: Use suitable exponentials.
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  2. #2
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    Quote Originally Posted by edgar davids View Post
    Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

    Edgar

    Let f be an entire function.
    (a) Prove that if e^f is bounded then f is constant.
    (b) Prove that if Re f is bounded then f is constant.
    Hint: Use suitable exponentials.
    What you need is Picard's little theorem, find it here.

    a) If e^{f(z)} is bounded then so is |f(x)|,
    let K>0 be a bound for |f(z)| then both:

    2Ke^{i\theta} and 3Ke^{i\theta}

    are not values taken by f(z), hence by Picard's theorem is a constant.


    b) Same as above let K>0 be a bound, then both:

    2K+i and 3K+i

    are not values taken by f(z), hence by Picard's theorem is a constant.

    RonL
    Last edited by CaptainBlack; November 14th 2006 at 06:21 AM.
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  3. #3
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    Liouvilles theorem

    This results follow easily from Liouville's theorem. An entire bounded function is contstant
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  4. #4
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    Quote Originally Posted by edgar davids View Post
    Anothre difficult question think im missing something here. If someonecould help me out would be much appreciated

    Edgar

    Let f be an entire function.
    (a) Prove that if e^f is bounded then f is constant.
    (b) Prove that if Re f is bounded then f is constant.
    Hint: Use suitable exponentials.
    Quote Originally Posted by petemarkou View Post
    This results follow easily from Liouville's theorem. An entire bounded function is contstant
    Liouville's theorem gives e^f is constant, and you need a little additional argument (since exp is not injective). For instance, 0=(e^f)'=f' e^f hence f'=0.

    And for b), use |e^{f}|=e^{\Re f} to reduce to the first question.
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    What you need is Picard's little theorem, find it here.

    a) If e^{f(z)} is bounded then so is |f(x)|
    There's probably no point in noting this (considering the delay), but anyway: a priori no, if e^f if bounded, then \Re f is bounded, but we don't know about f. If f is entire, then a posteriori it is constant hence bounded.
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