Complex variable - Maximum Modulus Principle

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Homework Statement



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).

Homework 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.

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?
 

Answers and Replies

  • #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?
 
  • #3
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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?
 
  • #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.
 
  • #5
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Thanks!
So..
Since f(z) is nonzero in R, g = 1/f is analytic in R.
Is it enough for proving g is analytic?
 
  • #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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