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    mobius transformation

    Find a Mobius transformation that maps the right half-plane to the upper half-plane carrying the point 7+5i to 3i.

    Can I get some help please?
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    Quote Originally Posted by dori1123 View Post
    Find a Möbius transformation that maps the right half-plane to the upper half-plane carrying the point 7+5i to 3i.
    This method is messy, but at least it ought to work. The idea is that you should have a library of known Möbius transformations that you combine in order to get the desired result.

    The transformation f(z) = \frac{z-1}{z+1} maps the right half-plane to the unit circle, and it takes 7+5i to \frac{73+10i}{89} (that's where the numbers start to get messy).

    The transformation g(z) = \frac{z-i}{z+i} maps the upper half-plane to the unit circle, and it takes 3i to 1/2. Its inverse transformation, from the unit disk to the upper half-plane, is given by g^{-1}(z) = \frac{i(1+z)}{1-z}.

    For any point a in the unit disk, the transformation h_a(z) = \frac{z-a}{1-\overline{a}z} maps the unit disk to itself, and takes a to the origin. Its inverse transformation is given by h_a^{-1}(z) = \frac{z+a}{1+\overline{a}z}, and it obviously takes the origin to a.

    Thus the composition of mappings g^{-1}\circ h_b^{-1}\circ h_a\circ f, where a = \tfrac{73+10i}{89} and b=1/2, will be a Möbius transformation that maps the right half-plane to the upper half-plane and carries the point 7+5i to 3i.

    Having written all that, I have a better idea. You can think of the right half-plane as consisting of all those points that are closer to 7+5i than they are to –7+5i. In other words, they are the points for which |z-7-5i| < |z+7-5i|. It follows that the Möbius transformation k(z) = \frac{z-7-5i}{z+7-5i} maps the right half-plane to the unit disk, and it takes 7+5i to the origin. So you can use that map to replace the composite map h_a\circ f in the above construction by the single map k, and that will avoid all the messy arithmetic.
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