Consider the 'lens' described by $\{z:|z-i|<\sqrt{2}\ \text{and}\ |z+i|<\sqrt{2} \}$ . We want to map this to the upper right quadrant using a Möbius transformation.
The two circles meet at $z=1,-1$ and so if we use the map $f:z \mapsto \frac{z-1}{z+1}$ we should send the lens to a sector centered on the origin.
I have two questions:
- How do we find the angle sweeped out by the sector? As $f$ is conformal, it will be equal to the angle at which the two circles meet; but how is this calculated? And is there some general method for working this out?
- After finding this angle, how do we work out how the sector is rotated relative to say the real axis after applying $f$? Again, is there some general way of doing this also?
After establishing this, the question becomes simple as we can use a power map to adjust to the correct angle and then a rotation to position the sector.
After responses to 1. and 2. I think my approach is sound, though I'd be interested in hearing if it isn't also.
The centers $\pm i$ of the circles and the circle intersection points $\pm1$ form a square, so the circles intersect at right angles (the sides of the square are tangents). Therefore, a power map is not needed here. (The general method of course involves some acrtan manipulations with these four points)
For the second question, the lens is symmetric with respect to the real axis and your transform leaves the real axis invariant, therefore the image is also symmetric with respect to the real axis. So you should rotate by $45^\circ$ in the end.