# Mobius Transformation

• Aug 5th 2009, 05:30 PM
josephtw
Mobius Transformation
Map the region between the two circles |z-1|=1, |z|=2 conformally to the upper-half plane (mapping 2 to Infinity).

I actually have an answer to this, but I'm looking for pointers on the process.

Thanks.
• Aug 6th 2009, 01:32 AM
Opalg
One way to construct conformal maps like this is to piece them together from simple building blocks. To do that, you need to have a "library" of known simple conformal maps. For example, you know that fractional linear maps take lines and circles to lines and circles. So a good place to start would be the map $\displaystyle z\mapsto\frac1{2-z}$. This maps 2 to infinity, and takes the circles |z–1|=1 and |z|=2 to the lines re(z)=1/4 and re(z)=1/2. So it maps the region R between the circles to the strip $\displaystyle S_1 = \{z:1/4<\text{re}(z)<1/2\}$.

Next, does our library of conformal maps include one that takes a strip to a half-plane? Yes it does! The exponential map takes the strip $\displaystyle S_2 = \{z:0<\text{im}(z)<\pi\}$ to the upper half-plane H.

So now all you need is an affine map to dilate and rotate $\displaystyle S_1$ to $\displaystyle S_2$, and then the composition of the three maps will take R to H. The result is the map $\displaystyle z\mapsto \exp\Bigl(i\Bigl(\frac{4\pi}{2-z}-\pi\Bigr)\Bigr)$. (Of course, that's not the unique answer, because there are many conformal maps from H to itself.)
• Aug 6th 2009, 06:54 AM
josephtw
Thank you, that confirms a lot things I thought were true. I guess one more detail I'd like to show is why 1/(2-z) maps the circles to Re z=1/4 and Re z=1/2. I got that result before by checking a few points.

Is there any better method to confirm this besides the following? We know that it contains infinity and therefore must be a line as opposed to a circle. We then check two points and then get the position and orientation of the line. (for each line) I'd like to work out an equation algebraically, but I guess I'm missing something.

Thanks.
• Aug 6th 2009, 10:48 AM
Opalg
Quote:

Originally Posted by josephtw
Thank you, that confirms a lot things I thought were true. I guess one more detail I'd like to show is why 1/(2-z) maps the circles to Re z=1/4 and Re z=1/2. I got that result before by checking a few points.

Is there any better method to confirm this besides the following? We know that it contains infinity and therefore must be a line as opposed to a circle. We then check two points and then get the position and orientation of the line.

That's exactly how I did it. It's a line, because it goes through the point at infinity, and you determine it by finding two points on it.

If you want to do it algebraically, you can argue like this. Suppose you want to find the image of the circle |z|=2 under the map $\displaystyle z\mapsto1/(2-z)$. Let $\displaystyle w=\frac1{2-z}$. Solve that equation for z, getting $\displaystyle z = \frac{2w-1}w$. If |z|=2 then |2w–1|=2|w|, or $\displaystyle |w-\tfrac12| = |w|$. That says that the distance from w to 1/2 is equal to the distance from w to 0. So w lies on the perpendicular bisector of those two points, which is the line re(w)=1/4.

Incidentally, there is something dubious about the wording of the original question. It asks for a conformal map that sends 2 to infinity. Of course, 2 is not in the (open) region R between the two circles. So what the question means is that if you take a path in R that approaches 2 then its image under the conformal map should go to infinity. But there are two ways that you can approach 2 in R: either from above the real axis, or from below it. If you use the conformal map that I suggested, and take a path that approaches 2 from above, then the image of the path approaches 0. But if you use a path that approaches 2 from below, then the image of the path goes to infinity. The point 2 has two images under any conformal map from R to H, and only one of these can be the point at infinity.