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 $f^{-1}:\\Delta\\to UHP$ . ∎

单词	UnitDiskUpperHalfPlaneConformalEquivalenceTheorem
释义	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 $f^{-1}:\\Delta\\to UHP$ . ∎
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