Möbius transformation composition

Question

Suppose we have

$F(z)=f(\phi(z))$ where $\phi$ is a Möbius transformation which maps points of the unit circle to points of the unit circle. Suppose also that there is an interval of length $\pi$ such that on that interval $f(e^{it})-f(e^{-it})>0$. How to prove then this is also true for $F$?

Here, I defined:

$\phi(z):=\frac{z_0-z}{1-\bar{z_0}z}$ with $z_0$ a fixed point of the unit disk (without the boundary) and $f$ a real harmonic continuous function defined on the unit disk. And $F(z)=f(\phi(z))$.

Anyone an idea please?

Answer 1

Möbius transformations preserve angles, not distances. That's called a conformal map. The Möbius transformations are the conformal automorphisms of the Riemann sphere (extended complex plane.) They map circles on the Riemann sphere to circles on the Riemann sphere. So viewed on the complex plane, they map a circle or line to a circle or line. For instance, the Cayley transform $\frac{z-i}{z+i}$ maps the real line bijectively to the unit circle. Obviously, the Euclidean distance is not preserved.

Answer 2

The statement is true if you use the "right" notion of distance.

Consider the upper half plane $\mathbb H := \{z\in \Bbb C: \operatorname{Im} z > 0\}$. For two points $z_1,z_2\in \mathbb H$ with $z_k = x_k + iy_k$ (for $k=1,2$), the hyperbolic distance $d(z_1,z_2)$ between them is defined as the infimum of $$\int_{t_1}^{t_2} \frac{\sqrt{x'(t)^2 + y'(t)^2}}{y}\, dt$$ taken over all paths $\gamma(t)$ with $\gamma(t_k) = z_k = x(t_k) + iy(t_k)$ for $k=1,2$.

Then, the following theorem holds.

Theorem. Möbius transformations with coefficients in $\Bbb R$ preserve hyperbolic lengths, i.e., if $\phi(z) = \frac{az+b}{cz+d}$ for $a,b,c,d\in \Bbb R$ and $z_1,z_2 \in \Bbb H$, then $d(z_1,z_2) = d(\phi(z_1), \phi(z_2))$.

For a proof, see these notes.