|
|
|
Viewing Version
2
of
'proof of conformal mapping theorem'
|
[ view 'proof of conformal mapping theorem'
|
back to history
]
| Title of object: |
proof of conformal mapping theorem |
| Canonical Name: |
ProofOfConformalMappingTheorem |
| Type: |
Proof |
| Created on: |
2003-07-23 18:51:32 |
| Modified on: |
2003-07-23 18:52:58 |
| Classification: |
msc:30C35 |
| Keywords: |
analytic function, conformal mapping |
Preamble:
% 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
|
Content:
Let $D\subset\mathbb{C}$ be a domain, and let $f:D\to\mathbb{C}$ 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)=\pmatrix{a & b \\ -b & a} = 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.
|
|
|
|
|
|
