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Finding Mobieus mapping

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Mobieus mapping that maps { z e C, |z| <= 1 } to a disk {z e C, |z - 1| <= 1} in a real axis.

    3. The attempt at a solution

    I have had an idea that Mobieus mapping is from C to C such that it is a homeomorfism and it has an inverse mapping.

    I am not sure how you can use it for the mapping.
     
  2. jcsd
  3. Sep 27, 2009 #2

    Dick

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    I'm not sure you do know what a mobius map is. You should probably look it up. It's a map of the form f(z)=(az+b)/(cz+d) with a, b, c and d complex constants. All you need is a translation.
     
  4. Oct 3, 2009 #3

    Your answer suggests me to

    1. to take the corners of the initial disk such as { (-1,0), (0,1), (1,0) } in x -axis
    2. to map them to the corner points { (1,0), (2,1), (3,0) } in the x-axis

    I can change the Mobieus mapping from
    (az + b) / (cz + d) to (z + b) / (cz + d)
    by canceling the "stabilising" variable "a".

    I should apparently have the z -coordinate in the mappings too.
    However, I do have three points which suggests me that I can find the Mobieus map:

    (-1, 0, 1) -> (1, 2, 3)

    I get the following mapping by calculating the equations
    M = z + 2,
    when b=2, c=0 and d=1.
     
  5. Oct 4, 2009 #4

    Dick

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    I don't understand what you are doing. What's wrong with f(z)=z+1? Unless I'm completely wrong about what the question is. You are just mapping one disk to another disk, right? What does 'in a real axis' mean?
     
  6. Oct 4, 2009 #5
    It seems to mean that I need to map the circle to a real line such that the inner points of the circle of the circle are above the real axis while the corner points are on the axis.

    f(z) = z + 1 seems to be too ok.
    Your mapping has the different selection of the corner points so you get different values for b, c and d.

    I get the mapping by first selecting the following corner points of the circle and then mapping them to (1,2,3) such that (-1, 0, 1) -> (1, 2, 3).

    This selection gives me the mobius mapping: f(z) = z + 2.
     
  7. Oct 4, 2009 #6

    Dick

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    Ok, so you want to map the unit disk to the upper half plane? f(z)=z+2 definitely doesn't work. It generally easier to construct these maps by combining a few basic ones. See what the image of unit disk is under the map g(z)=1/(1-z). It's maps the disk to a half-plane. It's not the half-plane you want, but can you change the mapping so that it is?
     
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