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Math Help - Find a Mobius transformation...

  1. #1
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    Find a Mobius transformation...

    Let D denote the unit disc |z|<1. Let the point A be inside D. Find a Mobius transformation which maps D bijectively onto itself and the point A to 0.

    I'm really not sure how to do this. Are you supposed to pick three points in D and map them to where you want them to go, eg. A -> 0, and put them into some formula? I know a Mobius transformation has the form
    f(z) = (az+b)/(cz+d),
    I just don't know how to find a,b,c,d.

    Thank you! x
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  2. #2
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    I think it's:

    T(z)=e^{i\theta}\frac{z-z_0}{1-\overline{z_0}z},\quad \theta\in [0,2\pi]
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  3. #3
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    Ah, now how on earth did you get that? Please explain the method! x
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  4. #4
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    Quote Originally Posted by charlie View Post
    Ah, now how on earth did you get that? Please explain the method! x
    It's a standard fractional linear transformation of the unit disc onto the unit disc. I don't know how to derive it though.
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  5. #5
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    A solutionH

    Here is a solution:

    start by noticing that in order to have f(z_o)=0, f is of the form
    f(z)=(z-zo)/cz+d).

    Now, in order to D to be preserved, the image of the unit circle C by the Mobius transform f must be preserved as well. (Note that when C is preserved, you have D that is preserved or is bijectively sent to the set |z|>1. Since A is inside the unit disc, it's the first possibility that prevails).

    Hence, for all $\theta$, |e^(i\theta)-z_o|=|ce^(i\theta)+d|.

    Compute this to get the system
    |c|^2+|d|^2=1+|z_o|^2
    conjugate(c).d=-z_o

    Put |c|^2=|zo|^2/|d|^2 in the first equation to get |d|=1.

    Finally, cz+d=-z.conjugate(z0)/conjugate(d)+d=d(-z.conjugate(zo)/|d|^2+1)=d(1-zconjugate(zo)), and
    f(z)=(1/d)*(z-zo)/(1-zconjugate zo), with |d|=1, which is the solution given above.

    I'm sure there must be a more elegant proof.
    Sorry for my awful english
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