# proof of conformal Möbius circle map theorem

Let $f$ be a conformal map from the unit disk $\mathrm{\Delta}$ onto itself. Let $a=f(0)$. Let ${g}_{a}(z)=\frac{z-a}{1-\overline{a}z}$. Then ${g}_{a}\circ f$ is a conformal map from $\mathrm{\Delta}$ onto itself, with ${g}_{a}\circ f(0)=0$. Therefore, by Schwarz’s Lemma for all $z\in \mathrm{\Delta}$ $|{g}_{a}\circ f(z)|\le |z|$.

Because $f$ is a conformal map onto $\mathrm{\Delta}$, ${f}^{-1}$ is also a conformal map of $\mathrm{\Delta}$ onto itself. ${({g}_{a}\circ f)}^{-1}(0)=0$ so that by Schwarz’s Lemma $|{({g}_{a}\circ f)}^{-1}(w)|\le |w|$ for all $w\in \mathrm{\Delta}$. Writing $w={g}_{a}\circ f(z)$ this becomes $|z|\le |{g}_{a}\circ f(z)|$.

Therefore, for all $z\in \mathrm{\Delta}$ $|{g}_{a}\circ f(z)|=|z|$. By Schwarz’s Lemma, ${g}_{a}\circ f$ is a rotation. Write ${g}_{a}\circ f(z)={e}^{i\theta}z$, or $f(z)={e}^{i\theta}{g}_{a}^{-1}$.

Therefore, $f$ is a Möbius Transformation^{}.

Title | proof of conformal Möbius circle map theorem |
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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 |