ProofOfConformalMappingTheorem 2003-07-23 18:51:32 2004-04-14 10:25:33 Proof conformal mapping theorem 0 analytic function conformal mapping % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 $$ (\tilde f\circ\gamma_1)'(0)= \frac{\partial(u,v)}{\partial(x,y)}(\gamma_1(0)) \cdot \frac{{\rm d}\gamma_1}{{\rm d}t}(0)= J(x,y)\begin{pmatrix} \frac{{\rm d}u_1}{{\rm d}t} \\ \frac{{\rm d}v_1}{{\rm d}t} \end{pmatrix} $$ and, similarly, $$ (\tilde f\circ\gamma_2)'(0)=J(x,y)\begin{pmatrix} \frac{{\rm d}u_2}{{\rm d}t} \\ \frac{{\rm d}v_2}{{\rm d}t} \end{pmatrix} $$ by the chain rule. We see that if $f'(z)\neq 0$, $f$ 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.