# Möbius map from circles to lines

I want to find a Möbius transformation that takes the two circles $C(i,3), C(-i,1)$ to the parallel lines $Re(z)=0, Re(z)=1$. I know that they intersect at -2i, which means I have to map $2i \mapsto \infty$. I'm not sure how to ensure that both get mapped to the appropriate line. I've tried using 2 other points on each circle to define a map, but it's never worked out quite right.

In situation like this, are there any canonical choices of points to consider other than the obvious, like intersections and such?

Also, if I want to show that any four concyclic points can be mapped to $\pm 1, \pm k, k \in (0,1)$, does it just come down to comparing the cross ratios?

## 2 Answers

About the two circles. Here is a plan: just send the point $-2i$ to $\infty$ somehow. For instance, by $z \to 1/(z+2i)$. Then look at what happens to the circles. They will map to two parallel lines, but probably not the lines that you want. But this is not a problem: you can simply add a second transformation that will map the two lines that you have to the two lines that you want.

As for the second question, it probably does. Have you tried it?

UPDATE: Here is a more detailed answer to the second question. We have four distinct concyclic points and we need to prove that there is a Möbius transform that sends them to $\pm1,\,\pm k$ where $k \in (0,\,1)$. Here is a rough sketch of a proof.

Step 1. The points are concyclic, so there is a Möbius transformation that maps them all onto the real axis. So we can assume from the start that our 4 points are all on the real axis.

Step 2. We have 4 points $z_1,\,z_2,\,z_3,\,z_4$ on the real axis. Prove that there exists such a $k\in (0,1)$ that two cross-ratios are equal: $$(z_1,z_2;z_3,z_4)=(k,-k;1,-1).$$ Technical detail: you might need to relabel points $z_i$ in a different order to do this.

Step 3. Now you can use the fact that Möbius transformations can be specified by three points. So there is a Möbius transformation $\omega$ such that $\omega(z_1)=k,\,\omega(z_2)=-k$ and $\omega(z_3)=1$. And you can prove that $\omega(z_4)$ will be automatically equal to $-1$ using the fact that $(z_1,z_2;z_3,z_4)=(k,-k;1,-1)$. This is it.

• I tried the second question using the cross ratio, but it doesn't tell me much, just that the ratio is real (which we already knew). I'm thinking now that I want to prove there is a line that that the four points are symmetric in pairs with respect to. Then if I map that line to the imaginary axis, I'll get symmetric points $\pm a, \pm b$ which I can scale down as appropriate. – Kannaguchi O. Nov 16 '12 at 23:22
• In general, such a line does not exist. If it existed then the 4 points would be vertices of a right trapezoid, which is not guaranteed. In this case it's probably easier to go with a more general approach and use the fact that Mobius trnasformations can be specified by three points. I've updated my answer with a brief explanation of the idea. – Dan Shved Nov 17 '12 at 7:58

Following up on Dan Shved's answer: I think you should choose the radius of the inversion circle around $-2i$ in such a way that the two lines end up having the desired distance between them. After that a rotation and translation will move them where you want.

[If after the inversion the lines have the wrong distance between them, I don't see how a rotation and/or translation can then map the lines where you want them.]

• Rotations and translations may be not enough, but we also have homotheties. Although applying a homothety after an inversion is almost the same as choosing an appropriate radius of the inversion circle in advance. – Dan Shved Nov 16 '12 at 16:52
• That's what I did. Before, I tried mapping the imaginary axis to the real axis and $-2 \mapsto \infty$, since this would be the desired configuration, by $4i \mapsto 1, 0 \mapsto 0$. This didn't work, though. Turns out I had to do $4i \mapsto 0, 0 \mapsto 1$, which, I guess, had the same orientation as the original circles. – Kannaguchi O. Nov 16 '12 at 16:56