# proof of conformal mapping theorem

Let $D\subset \u2102$ be a domain, and let $f:D\to \u2102$ be an
analytic function^{}. By identifying the complex plane $\u2102$ with
${\mathbb{R}}^{2}$, we can view $f$ as a function from ${\mathbb{R}}^{2}$ to
itself:

$$\stackrel{~}{f}(x,y):=(\mathrm{\Re}f(x+iy),\mathrm{\Im}f(x+iy))=(u(x,y),v(x,y))$$ |

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

$$J(x,y)=\frac{\partial (u,v)}{\partial (x,y)}=\left(\begin{array}{cc}\hfill {u}_{x}\hfill & \hfill {u}_{y}\hfill \\ \hfill {v}_{x}\hfill & \hfill {v}_{y}\hfill \end{array}\right).$$ |

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 $(r\mathrm{cos}\theta ,r\mathrm{sin}\theta )$
and get

$$J(x,y)=\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right)=r\left(\begin{array}{cc}\hfill \mathrm{cos}\theta \hfill & \hfill \mathrm{sin}\theta \hfill \\ \hfill -\mathrm{sin}\theta \hfill & \hfill \mathrm{cos}\theta \hfill \end{array}\right).$$ |

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 $\stackrel{~}{f}$ are $\stackrel{~}{f}\circ {\gamma}_{1}$ and $\stackrel{~}{f}\circ {\gamma}_{2}$, respectively, and their derivatives at $t=0$ are

$${(\stackrel{~}{f}\circ {\gamma}_{1})}^{\prime}(0)=\frac{\partial (u,v)}{\partial (x,y)}({\gamma}_{1}(0))\cdot \frac{\mathrm{d}{\gamma}_{1}}{\mathrm{d}t}(0)=J(x,y)\left(\begin{array}{c}\hfill \frac{\mathrm{d}{u}_{1}}{\mathrm{d}t}\hfill \\ \hfill \frac{\mathrm{d}{v}_{1}}{\mathrm{d}t}\hfill \end{array}\right)$$ |

and, similarly,

$${(\stackrel{~}{f}\circ {\gamma}_{2})}^{\prime}(0)=J(x,y)\left(\begin{array}{c}\hfill \frac{\mathrm{d}{u}_{2}}{\mathrm{d}t}\hfill \\ \hfill \frac{\mathrm{d}{v}_{2}}{\mathrm{d}t}\hfill \end{array}\right)$$ |

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

$$J/r=\left(\begin{array}{cc}\hfill \mathrm{cos}\theta \hfill & \hfill \mathrm{sin}\theta \hfill \\ \hfill -\mathrm{sin}\theta \hfill & \hfill \mathrm{cos}\theta \hfill \end{array}\right)$$ |

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 |