proof of conformal mapping theorem

Let D be a domain, and let f:D be an analytic functionMathworldPlanetmath. By identifying the complex plane with 2, we can view f as a function from 2 to itself:


with u and v real functions. The Jacobian matrix of f~ is


As an analytic function, f satisfies the Cauchy-Riemann equationsMathworldPlanetmath, so that ux=vy and uy=-vx. At a fixed point z=x+iyD, we can therefore define a=ux(x,y)=vy(x,y) and b=uy(x,y)=-vx(x,y). We write (a,b) in polar coordinates as (rcosθ,rsinθ) and get


Now we consider two smooth curves through (x,y), which we parametrize by γ1(t)=(u1(t),v1(t)) and γ2(t)=(u2(t),v2(t)). We can choose the parametrization such that γ1(0)=γ2(0)=z. The images of these curves under f~ are f~γ1 and f~γ2, respectively, and their derivatives at t=0 are


and, similarly,


by the chain ruleMathworldPlanetmath. We see that if f(z)0, f transforms the tangent vectorsMathworldPlanetmath to γ1 and γ2 at t=0 (and therefore in z) by the orthogonal matrixMathworldPlanetmath


and scales them by a factor of r. In particular, the transformation by an orthogonal matrix implies that the angle between the tangent vectors is preserved. Since the determinantMathworldPlanetmath of J/r is 1, the transformation also preserves orientation (the direction of the angle between the tangent vectors). We conclude that f is a conformal mappingMathworldPlanetmathPlanetmath at each point where its derivative is nonzero.

Title proof of conformal mapping theorem
Canonical name ProofOfConformalMappingTheorem
Date of creation 2013-03-22 13:47:26
Last modified on 2013-03-22 13:47:26
Owner pbruin (1001)
Last modified by pbruin (1001)
Numerical id 7
Author pbruin (1001)
Entry type Proof
Classification msc 30C35