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Conformal Mapping

  1. Mar 21, 2009 #1
    Let [itex]L:=\{z:|z-1|<1\} \cap \{z:|z-i|<1\}[/itex]. Find a Mobius transformation that maps L onto the sector [itex]\{z: 0< arg(z) < \alpha \}. What is the angle [itex]\alpha[/itex]?

    no idea of how about to set up the problem

    The intersection of the two circles forms a lens shaped region L with boundary curves, let's call them [itex]C_1[/itex] and [itex]C_2[/itex].

    i couldn't decide whether to write down a generic Mobius transformation [itex]f(z)=\frac{az+b}{cz+d}[/itex] and try and work with it (this would need me to define stuff like points that map to zero and infinity though would it not) or to use the fact that Mobius transformations are combinations of inversions, dilations, rotations and translations and try and sipmlify the Mobius transformation this way????

    i need some explanation....

    cheers
     
  2. jcsd
  3. Mar 21, 2009 #2
    The key idea is to break it down into simple pieces. The basic "moves" you can do with a mobius map are translation, inversion, dilation, and rotation. You can read more about them here:
    http://en.wikipedia.org/wiki/Mobius_transformation#Decomposition_and_elementary_properties

    You need to string together a series of "moves" that will take the lens shape and map it to a region between 2 parallel lines.

    The hard part, in my opinion, is how to map a circle to a line. The answer is to use the Mobius inversion f(z) = 1/z. Have a look at the following diagram from wikipedia to see what happens under an inversion (initial circle is blue, mapped "circle" is the green line, and the unit circle is red):
    Inversion_illustration2.png
     
  4. Mar 22, 2009 #3
    so 1/z will map the lens shaped region to a wedge then but how do we impose conditions such that one of the new lines lies along the positive real axis and the other lies alpha radians from the positive real axis?
     
  5. Mar 22, 2009 #4
    translation and rotation
     
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