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Version 3 |
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Let $D\subset\mathbb{C}$ be a domain, and let $f\colon D\to\mathbb{C}$ be an
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Let $D\subset\mathbb{C}$ be a domain, and let $f:D\to\mathbb{C}$ be an
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| analytic function. By identifying the complex plane $\mathbb{C}$ with |
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 |
$\mathbb{R}^2$, we can view $f$ as a function from $\mathbb{R}^2$ to |
| itself: |
itself: |
| $$ |
|
| \tilde f(x,y):=(\Re f(x+iy), \Im f(x+iy))=(u(x,y),v(x,y)) |
\tilde f(x,y):=(\Re f(x+iy), \Im f(x+iy))=(u(x,y),v(x,y)) |
| $$ |
|
| with $u$ and $v$ real functions. The Jacobian matrix of $\tilde f$ is |
with $u$ and $v$ real functions. The Jacobian matrix of $\tilde f$ is |
| $$ |
|
| J(x,y)=\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix} |
J(x,y)=\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix} |
| u_x & u_y \\ |
u_x & u_y \\ |
| v_x & v_y |
v_x & v_y |
| \end{pmatrix}. |
\end{pmatrix}. |
| $$ |
|
| As an analytic function, $f$ satisfies the Cauchy-Riemann equations, |
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 |
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 $b=u_y(x,y)=-v_x(x,y)$. |
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 $(r\cos\theta,r\sin\theta)$ |
We write $(a,b)$ in polar coordinates as $(r\cos\theta,r\sin\theta)$ |
| and get |
and get |
| $$ |
|
| J(x,y)=\begin{pmatrix} |
J(x,y)=\begin{pmatrix} |
| a & b \\ |
a & b \\ |
| -b & a |
-b & a |
| \end{pmatrix} = r\begin{pmatrix} |
\end{pmatrix} = r\begin{pmatrix} |
| \cos\theta & \sin\theta \\ |
\cos\theta & \sin\theta \\ |
| -\sin\theta & \cos\theta |
-\sin\theta & \cos\theta |
| \end{pmatrix}. |
\end{pmatrix} |
| $$ |
|
|
|
| Now we consider two smooth curves through $(x,y)$, which we |
Now we consider two smooth curves through $(x,y)$, which we |
| parametrize by $\gamma_1(t)=(u_1(t),v_1(t))$ and |
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 |
$\gamma_2(t)=(u_2(t),v_2(t))$. We can choose the parametrization such |
| that $\gamma_1(0)=\gamma_2(0)=z$. The images of these curves under |
that $\gamma_1(0)=\gamma_2(0)=z$. The images of these curves under |
| $\tilde f$ are $\tilde f\circ\gamma_1$ and $\tilde f\circ\gamma_2$, |
$\tilde f$ are $\tilde f\circ\gamma_1$ and $\tilde f\circ\gamma_2$, |
| respectively, and their derivatives at $t=0$ are |
respectively, and their derivatives at $t=0$ are |
| $$ |
|
| (\tilde f\circ\gamma_1)'(0)= |
(\tilde f\circ\gamma_1)'(0)= |
| \frac{\partial(u,v)}{\partial(x,y)}(\gamma_1(0)) |
\frac{\partial(u,v)}{\partial(x,y)}(\gamma_1(0)) |
| \cdot \frac{{\rm d}\gamma_1}{{\rm d}t}(0)= |
\cdot \frac{{\rm d}\gamma_1}{{\rm d}t}(0)= |
| J(x,y)\begin{pmatrix} |
J(x,y)\begin{pmatrix} |
| \frac{{\rm d}u_1}{{\rm d}t} \\ |
\frac{{\rm d}u_1}{{\rm d}t} \\ |
| \frac{{\rm d}v_1}{{\rm d}t} |
\frac{{\rm d}v_1}{{\rm d}t} |
| \end{pmatrix} |
\end{pmatrix} |
| $$ |
|
| and, similarly, |
and, similarly, |
| $$ |
|
| (\tilde f\circ\gamma_2)'(0)=J(x,y)\begin{pmatrix} |
(\tilde f\circ\gamma_2)'(0)=J(x,y)\begin{pmatrix} |
| \frac{{\rm d}u_2}{{\rm d}t} \\ |
\frac{{\rm d}u_2}{{\rm d}t} \\ |
| \frac{{\rm d}v_2}{{\rm d}t} |
\frac{{\rm d}v_2}{{\rm d}t} |
| \end{pmatrix} |
\end{pmatrix} |
| $$ |
|
| by the chain rule. We see that if $f'(z)\neq 0$, $f$ transforms the |
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 |
tangent vectors to $\gamma_1$ and $\gamma_2$ at $t=0$ (and therefore |
| in $z$) by the orthogonal matrix |
in $z$) by the orthogonal matrix |
| $$ |
|
| J/r=\begin{pmatrix} |
J/r=\begin{pmatrix} |
| \cos\theta & \sin\theta \\ |
\cos\theta & \sin\theta \\ |
| -\sin\theta & \cos\theta |
-\sin\theta & \cos\theta |
| \end{pmatrix} |
\end{pmatrix} |
| $$ |
|
| and scales them by a factor of $r$. In particular, the transformation |
and scales them by a factor of $r$. In particular, the transformation |
| by an orthogonal matrix implies that the angle between the tangent |
by an orthogonal matrix implies that the angle between the tangent |
| vectors is preserved. Since the determinant of $J/r$ is 1, the |
vectors is preserved. Since the determinant of $J/r$ is 1, the |
| transformation also preserves orientation (the direction of the angle |
transformation also preserves orientation (the direction of the angle |
| between the tangent vectors). We conclude that $f$ is a conformal |
between the tangent vectors). We conclude that $f$ is a conformal |
|
mapping at each point where its derivative is nonzero.
|
mapping.
|
