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Math Help - [SOLVED] Minimum of a complex function, proof

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Minimum of a complex function, proof

    I've no idea what theorem can be useful to prove the following : Let f be a continuous function in a closed and bounded region R, assume also that f is analytic and not constant in the interior of R.
    Assuming that f(z)\neq 0 in all R, prove that |f(z)| has a minimum and that it is reached in the frontier of R and never in the interior of R.
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  2. #2
    Super Member Matt Westwood's Avatar
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    I'm not sure, but I suspect this may be an application of Liouville's theorem:

    http://www.proofwiki.org/wiki/Liouvi...mplex_Analysis)

    ... check it out anyway, there may be techniques and insights in there which will give you some inspiration.
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  3. #3
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    If f(R)\ne 0, then what can you say about the (analytic) function g(R)=\frac{1}{f(R)} using the Maximum Modulus Principle?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by shawsend View Post
    If f(R)\ne 0, then what can you say about the (analytic) function g(R)=\frac{1}{f(R)} using the Maximum Modulus Principle?
    That g reaches a minimum in the frontier of R.
    Not sure I'm right. By the way I never heard this principle before. That's interesting, it reminds me of Extreme value theorem - Wikipedia, the free encyclopedia. Thanks a lot!
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  5. #5
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    By the Maximum Modulus Principle, g(z) reaches a maximum on the boundary which means that f(z) must reach a minimum on the boundary as well.
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