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Complex analysis, finding a bijection

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Let Omega = C\((-inf,-1]U[1,inf)), find a holomorphic bijection phi:eek:mega-->delta, where delta is the open unit disk


    2. Relevant equations
    Reimann Mapping Theorem
    Special Mapping formulas: can map wedges onto wedges, with deletion of real line from zero to infinity in order to deal with log.
    The unique bilinear mapping that sends 3 points onto; inf,0,1.

    3. The attempt at a solution
    so what this looks like, is all of C besides for the parts on the real line that goes from -inf to 1 and 1 to inf. also, if we consider C-hat and the point at infinity we can see that this is a simply connected domain and so by Reimann Mapping Theorem there exists a bijection between it and every other simply connected domain(of which delta is one). so, in its present form this domain is unworkable. so i want to somehow map it onto the real line from 0 to inf from which i can go from there. like i said, i know the unique bilinear mapping but that is for 3 discrete points, which does not work in this case since i have the whole line. also, i need it to be 1 to 1, since it must be a bijection and i already have a point at 1 and infinity.
     
  2. jcsd
  3. Apr 29, 2009 #2
    Your Omega makes me think of the function sin(z). Can you use it (suitably restricted) or its inverse (suitably restricted), composed with a few other mappings, to create your desired mapping?
     
  4. Apr 29, 2009 #3
    while interesting, the problem that i see with this is that the zeroes of sin(z) are points on the real line so it does not relate to my original omega. after a little discussion and searching we found a simplified form of the bilinear mapping when one of the points is -inf or +inf. however, this is still for 3 discrete points. and even if we say we map the interior as a point between our two ends i am a bit concerned that i might map the line on the left side of the plane onto the right half of the plane and that the point of infinity will then have two same points. one from the original omega and one from the mapping.
     
  5. Apr 29, 2009 #4
    while interesting, the problem that i see with this is that the zeroes of sin(z) are points on the real line so it does not relate to my original omega. after a little discussion and searching we found a simplified form of the bilinear mapping when one of the points is -inf or +inf. however, this is still for 3 discrete points. and even if we say we map the interior as a point between our two ends i am a bit concerned that i might map the line on the left side of the plane onto the right half of the plane and that the point of infinity will then have two same points. one from the original omega and one from the mapping.
     
  6. Apr 29, 2009 #5
    What is the image of the strip -pi/2 < x < pi/2 under the mapping f(z)=sin z?
     
  7. Apr 29, 2009 #6
    when you say strip you mean just that piece of the real line right? because we are only varying x. so it look like a standard sine function, oscillating between -1 and 1...which is the part of our real line that still exists in my omega. but i fail to see what the connection is between this fact and the composition of bijections that lead to the unit disk.
     
  8. Apr 29, 2009 #7
    No, I mean the infinite strip {x+iy : -pi/2 < x < pi/2}, between the two vertical lines x=-pi/2 and x=pi/2. It's easiest perhaps if you can just look up sin(z) in a table of conformal mappings.

    Remember, you may find bijections that go from the unit disc to Omega, instead of the other way around. One step at a time, transform something to Omega, or Omega to something. Then transform that something to another thing, or vice versa.
     
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