unit disk upper half plane conformal equivalence theorem

Theorem 1.

There is a conformal map from $\Delta$ , the unit disk, to $U H P$ , the upper half plane.

Proof.

Define $f\colon\hat{{\mathbb{C}}}\to\hat{{\mathbb{C}}}$ (where $\hat{{\mathbb{C}}}$ denotes the Riemann Sphere) to be $f(z)=\displaystyle{\frac{z-i}{z+i}}$ . Notice that $f^{-1}(w)=\displaystyle{i\frac{1+w}{1-w}}$ and that $f$ (and therefore $f^{-1}$ ) is a Mobius transformation.

Notice that $f(0)=-1$ , $f(1)=\displaystyle{\frac{1-i}{1+i}}=-i$ and $f(-1)=\displaystyle{\frac{-1-i}{-1+i}}=i$ . By the Mobius Circle Transformation Theorem, $f$ takes the real axis to the unit circle. Since $f(i)=0$ , $f$ maps $U H P$ to $\Delta$ and $f^{-1}:\Delta\to UHP$ . ∎

Title	unit disk upper half plane conformal equivalence theorem
Canonical name	UnitDiskUpperHalfPlaneConformalEquivalenceTheorem
Date of creation	2013-03-22 13:37:52
Last modified on	2013-03-22 13:37:52
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	12
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 30C20
Related topic	UnitDisk
Related topic	UpperHalfPlane
Related topic	MobiusTransformation
Related topic	MobiusCircleTransformationTheorem
Related topic	ConvertingBetweenThePoincareDiscModelAndTheUpperHalfPlaneModel