# Möbius maps and their fixed points

Why is it true that if a Möbius map, $f(z)$ fixes distinct $z_1,z_2\in \mathbb C_\infty$ that $f(z)$ either describes a rotation or has a pair of stable and unstable fixed points, such that iterations of $f(z)$ converges to the stable fixed point for $z\neq$ the unstable fixed point?

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The fairly standard argument is to conjugate with a Moebius transformation that sends $0$ to $z_1$ and $\infty$ to $z_2$. This gives you a Moebius transformation that fixes $0$ and $\infty$. A little algebra tells you this is just the map $z \longmapsto az$ for some complex number $a$.

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Thank you! How did you come up with it? – paul Feb 7 '12 at 19:57
Conjugation of a map is essentially the equivalent of a coordinate change. When you talk about "rotation about fixed points", that can be interpreted as "in certain coordinates the map looks like $z \longmapsto az$". So the type of your question basically leads one to think: conjugation. – Ryan Budney Feb 7 '12 at 20:01
Thanks, just curious what is the coordinate system we end up working in? – paul Feb 7 '12 at 20:26
I'm a little confused now. I'm not saying it is a coordinate change, but it's analogous. Just look through what one does in other context when one makes a coordinate change, like representing a matrix with respect to another basis, for example. That produces a conjugation. – Ryan Budney Feb 7 '12 at 20:49
Very helpful, thank you Ryan. – WishingFish Jul 25 '13 at 17:48