PlanetMath: proof of conformal mapping theorem

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proof of conformal mapping theorem

(Proof)

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 as a function from $\mathbb{R}^2$ to itself:

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

with

and

real functions. The Jacobian matrix of $\tilde f$ is

$\displaystyle J(x,y)=\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix} u_x & u_y \ v_x & v_y \end{pmatrix}.$

As an analytic function,

satisfies the Cauchy-Riemann equations, so that

and

. At a fixed point $z=x+iy\in D$ , we can therefore define

and

. We write

in polar coordinates as $(r\cos\theta,r\sin\theta)$ and get

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

Now we consider two smooth curves through , 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 are

$\displaystyle (\tilde f\circ\gamma_1)'(0)= \frac{\partial(u,v)}{\partial(x,y)}(... ...trix} \frac{{\rm d}u_1}{{\rm d}t} \ \frac{{\rm d}v_1}{{\rm d}t} \end{pmatrix}$