proof of conformal Möbius circle map theorem

Let f be a conformal map from the unit disk Δ onto itself. Let a=f(0). Let ga(z)=z-a1-a¯z. Then gaf is a conformal map from Δ onto itself, with gaf(0)=0. Therefore, by Schwarz’s Lemma for all zΔ |gaf(z)||z|.

Because f is a conformal map onto Δ, f-1 is also a conformal map of Δ onto itself. (gaf)-1(0)=0 so that by Schwarz’s Lemma |(gaf)-1(w)||w| for all wΔ. Writing w=gaf(z) this becomes |z||gaf(z)|.

Therefore, for all zΔ |gaf(z)|=|z|. By Schwarz’s Lemma, gaf is a rotation. Write gaf(z)=eiθz, or f(z)=eiθga-1.

Therefore, f is a Möbius TransformationMathworldPlanetmath.

Title proof of conformal Möbius circle map theorem
Canonical name ProofOfConformalMobiusCircleMapTheorem
Date of creation 2013-03-22 13:36:45
Last modified on 2013-03-22 13:36:45
Owner brianbirgen (2180)
Last modified by brianbirgen (2180)
Numerical id 5
Author brianbirgen (2180)
Entry type Proof
Classification msc 30E20
Related topic SchwarzLemma
Related topic MobiusTransformation
Related topic AutomorphismsOfUnitDisk