# Thread: mobius conjugate to az & z+1

1. ## mobius conjugate to az & z+1

how to prove this??

2. The first problem is false. Let $f(z) = (az+b)/(cz+d)$ be a Mobius transformation. Then $f^{-1}(z) = (dz - b)/(-cz+a)$. Simple calculations will show that $f\circ \mu z \circ f^{-1} = \mu z$ and $f\circ z+1 \circ f^{-1} = z+1$. Thus, $\mu z,z+1$ 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).

3. ## prob

Originally Posted by ThePerfectHacker
The first problem is false. Let $f(z) = (az+b)/(cz+d)$ be a Mobius transformation. Then $f^{-1}(z) = (dz - b)/(-cz+a)$. Simple calculations will show that $f\circ \mu z \circ f^{-1} = \mu z$ and $f\circ z+1 \circ f^{-1} = z+1$. Thus, $\mu z,z+1$ 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

4. Originally Posted by szpengchao
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
Maybe you misunderstand the book. If you tried the calculations I did above you will find that any conjugation with $\mu z$ will only give $\mu z$. Similarly with $z+1$. Thus, anything which is not either of these two, i.e. $1/z$ cannot be conjugate to either a rotated dilation or a translation.

5. ## book says

if f has two distinct fixed points a,b let g=(z-a)/(z-b),
then, gfg^(-1) fixes 0 and infinite.
so gfg^(-1) (z) = az

if f has one distinct point a, let g=1/(z-a)
then gfg^(-1)(z) = z+1

is it the right proof?

i am sure this argument is right.

6. is it the right proof?
The thing with two fixed points is correct. I made a mistake when I was doing a triple computation.

if f has one distinct point a, let g=1/(z-a)
then gfg^(-1)(z) = z+1
Why? I agree that $g\circ f \circ g^{-1}$ has $\infty$ as a fixed point. Also any translation $z+k,k\not =0$ has just one fixed point. Why you choose $k=1$?