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How does one transform the points outside of the circle in the z-plane, $|z-1/2|>1/2$, $x>0$, to a region between two parallel lines in the w-plane using the transformation $w(z)=\frac{az+b}{cz+d}$?

Edit1: enter image description here

Here is a picture of the region. This is the problem:

Find a linear fractional transformation w = f(z) that transforms the shaded region to an infinite vertical strip in the w-plane. Sketch the mapping.

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    $\begingroup$ I don't think that's possible $\endgroup$
    – mercio
    Apr 29, 2017 at 17:31
  • $\begingroup$ @DanielFischer: A vertical strip is the intersection of two half planes, so your observation does not prove it impossible. $\endgroup$
    – user14972
    Apr 30, 2017 at 2:37
  • $\begingroup$ $w=\frac{1}{z}$ maps $\{z\mid |z-1/2|>1/2, \operatorname{Re}z >0\}$ onto $\{w\mid 0<\operatorname{Re} w<1\}$. $\endgroup$
    – ts375_zk26
    Apr 30, 2017 at 5:44
  • $\begingroup$ @Hurkyl I overlooked the "$x > 0$" part. Taking that into account, we have a region bounded by two tangent circles, and a parallel strip is of the same form. $\endgroup$ Apr 30, 2017 at 13:39

1 Answer 1

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Linear fractional transformations are determined by their values at three points. Furthermore, lines and circles get transformed to lines and circles.

The trick to problems like this is:

Choose three points for the output that ensure the end result has the shape you want
Choose an appropriate three points from the input
Construct the linear fractional transformation that has that effect on those three points.

In the input space, you have two key lines and circles. To get the output shape you want:
You want both of them to map to lines — so $\infty$ has to lie on both of their images
You want one to be vertical — so pick two points with the same $x$-coordinate and insist they lie on one of the images.

Note that your two shapes only intersect in one point — so after transformation they will still only intersect in one point. If that point of intersection is $\infty$, you're guaranteed that they are parallel lines.

So, if you ensure the output has this configuration of three points, you're guaranteed that the output of the transformation will be a pair of vertical lines. You just need to play with the orientation (e.g. swap the two points chosen in the second bullet) to toggle whether or not the shaded region gets mapped to the interior or the exterior of the strip.

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