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Cauchy-Riemann方程

设$f:\mathbf{C}\to \mathbf{C}$是一个复函数.我们说$f$在点$z_0\in \mathbf{C}$处复可微,当且仅当存在复数$f’(z_0)$,使得
\begin{equation}
\label{eq:1}
f(z)-f(z_0)=f’(z_0)(z-z_0)+o(z-z_0),
\end{equation}
其中$\lim_{z\to z_0}\frac{o(z-z_0)}{z-z_0}=0$.称$f’(z_0)$为函数$f$在点$z_0$处的导数.

记$z=x+iy$,$z_0=x_0+iy_0$,$f(z)=u(x,y)+iv(x,y)$,$f’(z_0)=a+ib$.其中$u(x,y),v(x,y)$为从$\mathbf{R}^2$到$\mathbf{R}$的函数,$x,y,x_0,y_0,a,b\in \mathbf{R}$.则式\eqref{eq:1}等价于
\begin{equation}
\label{eq:2}
\begin{pmatrix}
u(x,y)-u(x_0,y_0)\\
v(x,y)-v(x_0,y_0)
\end{pmatrix}=
\begin{pmatrix}
a&-b\\
b&a
\end{pmatrix}
\begin{pmatrix}
x-x_0\\
y-y_0
\end{pmatrix}+
\begin{pmatrix}
o_1(x-x_0,y-y_0)\\
o_2(x-x_0,y-y_0)
\end{pmatrix},
\end{equation}
其中$o_1(x-x_0,y-y_0)+io_2(x-x_0,y-y_0)=o((x-x_0)+i(y-y_0))$.式\eqref{eq:2}说明,从$\mathbf{R}^2$到$\mathbf{R}^2$的函数$F:(x,y)\to (u(x,y),v(x,y))$是可微的,且
$$
\begin{pmatrix}
\frac{\partial u}{\partial x}(x_0,y_0)&\frac{\partial u}{\partial y}(x_0,y_{0})\\
\frac{\partial v}{\partial x}(x_{0},y_{0})&\frac{\partial v}{\partial y}(x_0,y_0)
\end{pmatrix}=
\begin{pmatrix}
a&-b\\
b&a
\end{pmatrix}.
$$
可见,
\begin{equation}
\label{eq:3}
\frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial
y}(x_0,y_0),\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}(x_0,y_0).
\end{equation}
这就是Cauchy-Riemann方程.

反之,如果$F:(x,y)\to (u(x,y),v(x,y))$在点$(x_0,y_0)$可微,且偏导数满足Cauchy-Riemann方程\eqref{eq:3},则$f$在$z_0$处复可微.