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Finding Mobius Transformation

  1. Feb 9, 2012 #1
    I'm looking for the error in my understanding here, not help with the problem itself. I'm making some kind of mistake, so I've listed out everything I think I know, and I'm hoping someone can either tell me what I'm misunderstanding, or tell me there is an error in the problem statement:

    Problem Statement: find the Mobius transformation taking the circle |z|=1 to |z+2|=1 such that T(-1)=-3 and T(i)=-1.

    What I think:

    *Both the circle and its image have radius 1, with the first centered at the origin and the second at -2.

    *The additional points we are given mappings for are -1 and i, which are the endpoints to an arc of a quarter circle. These are mapping to -3 and -1 which are the bounds for the half circle. This implies the mapping is going around twice for once around the circle being mapped, which would imply it is not a bijection, and Mobius transformations are bijections.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2012 #2

    Dick

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    Sure you can do it. You can map any three points in the complex plane to any other three points with a Mobius transformation, right? Just pick a third point on each of the two circles and that will fix a transformation. I haven't tried to visualize it in terms of dilations, rotations, etc.
     
  4. Feb 9, 2012 #3
    Yes about the three points. My problem isn't so much with that, I'm okay calculating it, I'm just bothered by how it can work.

    If I translate a circle, rotate a circle, or dilate a circle, or invert a circle, I don't see how points on a quarter arc could move to a half arc. Does my visualization problem make sense? I have to be understanding something incorrectly. Maybe inversion since the other three are so simple to visualize.

    I tried finding examples, but in every example, the points stay in the same relative (for lack of a better word) location
     
  5. Feb 9, 2012 #4

    Dick

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    Yes, it's inversions. Replace |z+2|=1 with, say |z+100|=99. It should be visually clear that most of the points on the circle map to points near 0. z=(-1) and points nearby don't. It stretches angles on the circle. It's a good thing to visualize, keep it up!
     
    Last edited: Feb 9, 2012
  6. Feb 9, 2012 #5
    Thanks!!
     
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