# unit disk upper half plane conformal equivalence theorem

###### Theorem 1.

There is a conformal map from $\Delta$, the unit disk, to $UHP$, 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 $UHP$ to $\Delta$ and $f^{-1}:\Delta\to UHP$. ∎

Title unit disk upper half plane conformal equivalence theorem UnitDiskUpperHalfPlaneConformalEquivalenceTheorem 2013-03-22 13:37:52 2013-03-22 13:37:52 CWoo (3771) CWoo (3771) 12 CWoo (3771) Theorem msc 30C20 UnitDisk UpperHalfPlane MobiusTransformation MobiusCircleTransformationTheorem ConvertingBetweenThePoincareDiscModelAndTheUpperHalfPlaneModel