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Is there a Möbius transformation mapping the upper half plane onto itself that interchange two preassigned points in the upper half plane? If so , how many such Möbius transformations are there?

Here are steps that I followed:

1.we know we can transform upper half plane onto unit circle.

  1. I want to find a Möbius transformation that maps unit circle to unit circle but interchanging the preassigned points.

3. Finally , if I find a transformation that maps the unit circle to the upper half plane.Then I think we are done.

I have problem in finding the Möbius transformation that maps unit circle to unit circle with interchanging property that already mention.

Does someone have any idea about this?

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    $\begingroup$ the transformations that take the upper half plane to itself are easier to describe, and work with, than what you suggest $\endgroup$ – Will Jagy Oct 1 '14 at 4:12
  • $\begingroup$ Question with a related answer $\endgroup$ – ccorn Oct 1 '14 at 5:56
  • $\begingroup$ What I think , we need three transformation to do this. But , I am having problem to see how to interchange the points $\endgroup$ – user178061 Oct 2 '14 at 16:36
  • $\begingroup$ You know that you can move the problem to the unit disk. I think that can indeed simplify matters somewhat. Can you show that there exists an automorphism of the unit disk that interchanges $0$ and any preassigned point $a$ in the unit disk? Can you then show that there is an automorphism of the unit disk interchanging two arbitrary preassigned $a,b$ in the unit disk? $\endgroup$ – Daniel Fischer Oct 2 '14 at 21:16
  • $\begingroup$ It makes sense. This is what I was trying to do but I have the problem to show there is an automorphism of the unit disk interchanging two arbitrary preassigned points. $\endgroup$ – user178061 Oct 3 '14 at 3:08
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A Möbius transformation that maps the upper half plane to itself must map the boundary, i.e. the real line, to itself as well. With the standard definition for the transformation, $$f(z) = \frac{az+b}{cz+d},$$ it is fairly straightforward to see that the parameters $a, b, c, d$ must be real to map the real line to itself (try plugging in $z=0$, $z=\infty$, and then one other real number).

Then since $f$ is supposed to interchange $z_1$ and $z_2$, that means applying $f$ twice should send $z_1$ back to itself, and the same for $z_2$. Note that $f(f(z))$ is again another Möbius transformation with real coefficients. You can now solve the equation for the fixed points of the transformation (or look up the formula). This should answer the question about whether the transformation is possible.

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  • $\begingroup$ I did not get how applying f twice takes the points back to itself and how does this interchanges the points? $\endgroup$ – user178061 Oct 14 '14 at 3:51

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