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Complex variable - Maximum Modulus Principle

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let a function f be continuous on a closed bounded region R, and let it be analytic and not constant throughout the interior of R. Assuming that f(z) != 0 anywhere in R, prove that |f(z)| has a minimum value m in R which occurs on the boundary of R and never in the interior. Do this by applying the corresponding result for maximum values to the function g(z) = 1/f(z).

    2. Relevant equations
    Corollary. Suppose that a function f is continuous on a closed bounded region R and that it is analytic and not constant in the interior of R. Then the maximum value of |f(z)| in R, which is always reached, occurs smoewhere on the boundary of R and never in the interior.

    3. The attempt at a solution
    Suppose that g = 1/f is continuous on a closed bounded region R and that it is analytic and not constant in the interior of R. By Corollary, |g(z)| has a maximum value, and
    when |g(z)| has the maximum value, |1/g(z)|=|f(z)| has the minimum value. Complete.

    Is it right approach?
     
  2. jcsd
  3. Oct 19, 2009 #2

    Dick

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    Basically. But you are skimming over an issue. You can't assume g=1/f is analytic. You are only given that f is analytic. You have to prove g is analytic. What might this have to do with the assumption that f(z)!=0 in R?
     
  4. Oct 19, 2009 #3
    Thanks,
    I think
    since f(z) is not equal to zero, 1/f(z) is entire functino and also anlytic.
    Thus I can assume g is analytic...
    How about this idea?
     
  5. Oct 19, 2009 #4

    Dick

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    f(z) is only assumed to be nonzero in R. That doesn't make it 'entire', but yes, that's the missing ingredient. g is analytic in R.
     
  6. Oct 19, 2009 #5
    Thanks!
    So..
    Since f(z) is nonzero in R, g = 1/f is analytic in R.
    Is it enough for proving g is analytic?
     
  7. Oct 19, 2009 #6

    Dick

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    You've probably already proved that, haven't you? If a(z)=b(z)/c(z) and c(z0) is not zero, then a(z) is analytic at z=z0. Right?
     
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