Basic Complex Analysis: Maximum Modulus?

  • Thread starter snipez90
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  • #1
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Homework Statement


Let f and g be two holomorphic functions in the unit disc D1 = {z : |z| < 1}, continuous in D1, which do not vanish for any value of z in the closure of D1. Assume that |f(z)| = |g(z)| for every z in the boundary of D1 and moreover f(1) = g(1). Prove that f and g are the same function.


Homework Equations


Maximum modulus?


The Attempt at a Solution


Last one for now.

Okay I had enough sense to gather from the hint that f and g don't vanish that I should define h = f/g. Then h(1) = 1, |h(z)| = 1 for z on the boundary of D1. Now I feel like maximum modulus will give that h(z) = 1 on D1, but I'm making a logical leap. Can someone help me out here? Thanks.
 

Answers and Replies

  • #2
Dick
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Since |h(z)|=1 on the boundary, what does that tell you about |h(z)| in the interior? Since |h(z)| then achieves its maximum in the interior, what does that tell you about h(z)? Now use h(1)=1.
 
  • #3
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By maximum modulus, [itex]|h(z)| \leq 1[/itex] in the interior. But why does |h(z)| achieve a maximum in the interior?
 
  • #4
Dick
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There's an analogous minimum modulus theorem as well isn't there? Since h(z) doesn't vanish, replace h(z) with 1/h(z).
 
  • #5
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Gotcha, I considered 1/|h| and things came together, but yes that does give you minimum modulus I guess. Thanks.

If you don't mind could you take a look at

https://www.physicsforums.com/showthread.php?t=484067

to spot check the proof? The method I chose seemed like the easiest way to estimate the derivatives.

Also that Frobenius eigenvalue problem that I showed you awhile back had a deeper solution than the method we agreed on (which was like raising the linear operator to the 7th power). I think I also made a miscalculation but I might post again if I get around to thinking about it again. Of course nothing of the sort actually showed up on the final.
 
  • #6
Dick
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I don't remember any Frobenius eigenvalue thing. Can you remind me?
 

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