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Composing functions that are conformal

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    let f and g be conformal analytic functions with the range of f a subset of the domain of g. Show that the composition g(f(z)) is also conformal.


    2. Relevant equations





    3. The attempt at a solution

    Well I know that if the function is conformal it is 1-1, analytic, and its derivative is non zero. So i'm thinking that I would need to show that since f'(z)≠0 and g'(z)≠0 then (g(f(z))'≠0 and then show that since f and g are analytic, g(f(z)) is also analytic. Then the composition will be both analytic and its derivative will not be zero so it will be conformal. Am I working in the right direction?
     
  2. jcsd
  3. Apr 8, 2012 #2

    Office_Shredder

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    That's one way to do it. The other way is to just go straight with the definition and show that g(f(z)) preserves angles
     
  4. Apr 8, 2012 #3
    Is it possible to show, in general, that g(f(z)) preserves angles? I thought we would have to have actual functions defined for that..
     
  5. Apr 8, 2012 #4
    How would I prove the derivative part of this? I know that (g(f(z)))'=g'(f(z))*f'(z) and obviously f' is not zero but i'm not sure how to show with certainty that g'(f(z)) is not zero...
     
  6. Apr 8, 2012 #5

    Dick

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    You are given that "the range of f a subset of the domain of g".
     
  7. Apr 8, 2012 #6

    HallsofIvy

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    Let P, Q, and R be three points such that angle PQR has measure [itex]\theta[/itex]. Then, since f is conformal, P'=f(P), Q'=f(Q), and R'= f(R) are three points such that angle P'Q'R' has measure [itex]\theta[/itex]. And, since g is conformal, P''= g(P'), Q''= g(Q'), and R''= g(R') are three points such that ....
     
  8. Apr 8, 2012 #7
    Such that the angle P''Q''R'' is θ meaning that the composition is angle preserving. right?
     
  9. Apr 8, 2012 #8
    can we say that since the range f(z) is in the domain of g then g(f(z)) is basically g(z)? and then it is non zero from there?
     
  10. Apr 8, 2012 #9

    Dick

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    f being conformal doesn't imply that. The 'angle preserving' in conformal refers to the angles between intersecting curves. Not to angles between points. That only works if f maps lines to lines.
     
  11. Apr 8, 2012 #10

    Dick

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    g' is nonzero on its domain. f(z) is in that domain. I wouldn't say 'g(f(z)) is basically g(z)', that's not very true.
     
  12. Apr 8, 2012 #11
    well when we say g(z), that z includes f(z) since the range of f is a subset of the domain of g... ie. g(z) takes values from its domain, which includes f(z).. right?
     
  13. Apr 8, 2012 #12

    Dick

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    Right. Just say f(z) is in the domain of g. Don't say g(f(z)) is the same as g(z).
     
  14. Apr 8, 2012 #13
    ok, got it. Thanks for your help
     
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