设 $f:\ \mathbb{R}^2\rightarrow\mathbb{R}^2$ 是 $C^\infty$ 映射($f(x,y)=\bigl(f_1(x,y),f_2(x,y)\bigr)$), 且满足 Cauchy-Riemann 方程
\[
\begin{cases}
\frac{\partial f_1}{\partial x}=\frac{\partial f_2}{\partial y}\\
\frac{\partial f_1}{\partial y}=-\frac{\partial f_2}{\partial x}
\end{cases}
\]
证明:
(1) $Df(x,y)=0$ 当且仅当 $\det(Df(x,y))=0$.
(2) $f:\ \mathbb{R}^2\rightarrow\mathbb{R}^2$ 的局部逆(如果存在), 也满足 Cauchy-Riemann 方程.
(3) 若 $f$ 不满足 Cauchy-Riemann 方程, 则 (1) 不正确, 请给出反例.
1
(1)
\[Df(x,y)=
\begin{pmatrix}
\frac{\partial f_1}{\partial x}&\frac{\partial f_1}{\partial y}\\
\frac{\partial f_2}{\partial x}&\frac{\partial f_2}{\partial y}
\end{pmatrix}=
\begin{pmatrix}
\frac{\partial f_1}{\partial x}&-\frac{\partial f_2}{\partial x}\\
\frac{\partial f_2}{\partial x}&\frac{\partial f_1}{\partial x}
\end{pmatrix}
\]
\[\det(Df(x,y))=\biggl(\frac{\partial f_1}{\partial x}\biggr)^2+\biggl(\frac{\partial f_2}{\partial x}\biggr)^2=0\Leftrightarrow\frac{\partial f_1}{\partial x}=0=\frac{\partial f_2}{\partial x}\Leftrightarrow Df(x,y)=0.\]
(2)
若 $f:\ \mathbb{R}^2\rightarrow\mathbb{R}^2$ 在某点附近存在逆映射 $f^{-1}:\ \mathbb{R}^2\rightarrow\mathbb{R}^2$, $f^{-1}(u,v)=(x,y)$. 则由逆映射定理,
\[Df^{-1}(u,v)=\bigl(Df(x,y)\bigr)^{-1}=\frac{1}{\det(Df(x,y))}
\begin{pmatrix}
\frac{\partial f_1}{\partial x}&\frac{\partial f_2}{\partial x}\\
-\frac{\partial f_2}{\partial x}&\frac{\partial f_1}{\partial x}
\end{pmatrix}
\]
因此, $f^{-1}$ 在该点附近也满足 Cauchy-Riemann 方程. 具体的, 若记 $d=\det(Df(x,y))$, 则
\[
\frac{\partial f^{-1}_1}{\partial u}=\frac{1}{d}\cdot\frac{\partial f_1}{\partial x}=\frac{\partial f^{-1}_2}{\partial v}
\]
\[
\frac{\partial f^{-1}_1}{\partial v}=\frac{1}{d}\cdot\frac{\partial f_2}{\partial x}=-\frac{\partial f^{-1}_2}{\partial u}
\]
(3)
反例 $f(x,y)=(\sin(xy),\cos(xy))$, 显然不满足 Cauchy-Riemann 方程.
\[
Df(x,y)=
\begin{pmatrix}
y\cos(xy) & x\cos(xy)\\
-y\sin(xy) & -x\sin(xy)
\end{pmatrix}
\]
但是
\[\det(Df(x,y))=-xy\sin(xy)\cos(xy)+xy\sin(xy)\cos(xy)=0\]