how to prove this??
The first problem is false. Let be a Mobius transformation. Then . Simple calculations will show that and . Thus, are invariant under conjugation and belong to the center of the Mobius group. This means that they are only conjugate to themselves. Therefore the problem is wrong (except for the part which says dilations and translations are not conjugate).
i saw this as a theorem on a textbook. it says,
if a mobius transformation f has exactly two fixed points, then it is conjugate to some map az. If f has exactly one fixed point, it is conjugate to z+1
and there is no proof.
but it shows any mobius map can be written as matrix, and can diagonalize another matrix A
The thing with two fixed points is correct. I made a mistake when I was doing a triple computation.is it the right proof?
Why? I agree that has as a fixed point. Also any translation has just one fixed point. Why you choose ?if f has one distinct point a, let g=1/(z-a)
then gfg^(-1)(z) = z+1