proof of conformal mapping theorem

Let $D\subset\mathbb{C}$ be a domain, and let $f\colon D\to\mathbb{C}$ be an analytic function. By identifying the complex plane $\mathbb{C}$ with $\mathbb{R}^{2}$ , we can view $f$ as a function from $\mathbb{R}^{2}$ to itself:

$\tilde{f}(x,y):=(\Re f(x+iy),\Im f(x+iy))=(u(x,y),v(x,y))$

with $u$ and $v$ real functions. The Jacobian matrix of $\tilde{f}$ is

$J(x,y)=\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix}u_{x}&u_{y}\\ v_{x}&v_{y}\end{pmatrix}.$

As an analytic function, $f$ satisfies the Cauchy-Riemann equations, so that $u_{x}=v_{y}$ and $u_{y}=-v_{x}$ . At a fixed point $z=x+iy\in D$ , we can therefore define $a=u_{x}(x,y)=v_{y}(x,y)$ and $b=u_{y}(x,y)=-v_{x}(x,y)$ . We write $(a,b)$ in polar coordinates as $(r\cos\theta,r\sin\theta)$ and get

$J(x,y)=\begin{pmatrix}a&b\\ -b&a\end{pmatrix}=r\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}.$

Now we consider two smooth curves through $(x,y)$ , which we parametrize by $\gamma_{1}(t)=(u_{1}(t),v_{1}(t))$ and $\gamma_{2}(t)=(u_{2}(t),v_{2}(t))$ . We can choose the parametrization such that $\gamma_{1}(0)=\gamma_{2}(0)=z$ . The images of these curves under $\tilde{f}$ are $\tilde{f}\circ\gamma_{1}$ and $\tilde{f}\circ\gamma_{2}$ , respectively, and their derivatives at $t=0$ are

$(\tilde{f}\circ\gamma_{1})^{\prime}(0)=\frac{\partial(u,v)}{\partial(x,y)}(% \gamma_{1}(0))\cdot\frac{{\rm d}\gamma_{1}}{{\rm d}t}(0)=J(x,y)\begin{pmatrix}% \frac{{\rm d}u_{1}}{{\rm d}t}\\ \frac{{\rm d}v_{1}}{{\rm d}t}\end{pmatrix}$

and, similarly,

$(\tilde{f}\circ\gamma_{2})^{\prime}(0)=J(x,y)\begin{pmatrix}\frac{{\rm d}u_{2}% }{{\rm d}t}\\ \frac{{\rm d}v_{2}}{{\rm d}t}\end{pmatrix}$

by the chain rule. We see that if $f^{\prime}(z)\neq 0$ , $f$ transforms the tangent vectors to $\gamma_{1}$ and $\gamma_{2}$ at $t=0$ (and therefore in $z$ ) by the orthogonal matrix

$J/r=\begin{pmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}$

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 determinant 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 mapping 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