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Complex Mapping - Is transforming boundaries enough?

  1. Sep 16, 2015 #1
    Say I will make the transformation from the ##z## plane to the ##w## plane. Moreover, I'll transform a region ##R## with boundary ##C## in the ##z## plane to something in the ##w## plane.

    Why is it that if I know the equations for ##C## then I can transform these and immediately know that ##R## will be inside the transformed boundaries? Why isn't it the case that some point in ##R## maps into some point not inside ##C'## (which is my transformed boundary)?

    Thanks.
     
  2. jcsd
  3. Sep 16, 2015 #2

    RUber

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    You have to say something about the transformation. Normally, you are using linear, or at least continuous mappings which by definition will preserve this inside-outside relationship.
     
  4. Sep 16, 2015 #3
    According to what you say I think continuous mappings would be what I'm dealing with. Strangely I didn't think of continuity, which makes this more intuitive, but not obvious to me.

    I see that every point in the neighborhood of ##z_o## in ##R## will also be mapped in ##R'## as a simple closed region. I guess I can visualize these small neighborhoods expanding until they reach the transformed contour...

    Is this the argument?

    I was told this was because analytic transformations are "open maps" is this what they mean by it?
     
  5. Sep 16, 2015 #4

    Svein

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    Counterexample: Let [itex] C= (z:\lvert z \lvert = 1)[/itex], then [itex]R= (z:\lvert z \lvert < 1) [/itex]. Now transform this using [itex] w=\frac{1}{z}[/itex]. This maps C onto C, but R is mapped to the outside of C.
     
  6. Sep 16, 2015 #5
    Svein, but if the function ##w## is analytic in every point in ##R## then what I say will hold right?
     
  7. Sep 16, 2015 #6

    Svein

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    Sorry, I do not know. I checked out Ahlfors, and he carefully does not make that statement. It depends very much on C and the concept of inside.
     
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