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柯西-黎曼方程

1. **点可导的充要条件**

定理 函数 $w=f(z)=u(x, y)+i v(x, y)$ 在点 $z=x+i y$ 处可导  的充要条件是: $u(x, y)$ 和 $v(x, y)$ 在点 $(x, y)$ 处可微, 且满足柯西一黎曼 (Cauchy-Riemann ) 方程:

$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} . \quad \text { (简称 } C-R \text { 方程) }
$$

附 实二元函数 $u(x, y)$ 可微的含义:

$$
\begin{aligned}
\Delta u & =A \Delta x+B \Delta y+o\left(\sqrt{\Delta x^2+\Delta y^2}\right) \\
& =\frac{\partial u}{\partial x} \Delta x+\frac{\partial u}{\partial y} \Delta y+o\left(\sqrt{\Delta x^2+\Delta y^2}\right)
\end{aligned}
$$

**证明：**

1. 点可导的充要条件

**必要性 **“ $\Rightarrow "$ 若 $w=f(z)=u+i v$ 在 $z=x+i y$ 处可导,则必可微, 即 $\Delta w=f^{\prime}(x) \Delta z+o(|\Delta z|)$,记 $f^{\prime}(z)=a+i b$, 由 $\Delta w=\Delta u+i \Delta v, \Delta z=\Delta x+i \Delta y$ 有

$$
\Rightarrow \begin{aligned}
\Delta u+i \Delta v=(a+b i)(\Delta x+i \Delta y)+o(|\Delta z|), & \left\{\begin{array}{l}
\Delta u=a \Delta x-b \Delta y+o(|\Delta z|), \\
\Delta v=b \Delta x+a \Delta y+o(|\Delta z|),
\end{array}\right.
\end{aligned}
$$

故 $u(x, y)$ 和 $v(x, y)$ 在点 $(x, y)$ 处可微, 且

$$
a=\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad-b=\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
$$

**充分性** “" $\Leftarrow "$ 若 $u(x, y)$ 和 $v(x, y)$ 在点 $(x, y)$ 处可微, 则

$$
\left\{\begin{array}{l}
\Delta u=u_x^{\prime} \Delta x+u_y^{\prime} \Delta y+o(|\Delta z|), \\
\Delta v=v_y^{\prime} \Delta y+v_x^{\prime} \Delta x+o(|\Delta z|),
\end{array}\right.
$$

又由 $u$ 和 $v$ 满足 $C-R$ 方程: $u_x^{\prime}=v_y^{\prime}, u_y^{\prime}=-v_x^{\prime}$, 得

$$
\begin{aligned}
&\left\{\begin{array}{l}
\Delta u=u_x^{\prime} \Delta x-v_x^{\prime} \Delta y+o(|\Delta z|), \\
\Delta v=u_x^{\prime} \Delta y+v_x^{\prime} \Delta x+o(|\Delta z|),
\end{array}\right. \\
& \Rightarrow \Delta w=\Delta u+i \Delta v=\left(u_x^{\prime}+i v_x\right)(\Delta x+i \Delta y)+o(|\Delta z|),
\end{aligned}
$$

即 $f(z)$ 在 $z=x+i y$ 处可微 (可导), 且 $f^{\prime}(z)=u_x^{\prime}+i v_x^{\prime}$.

**求导公式**

若 $f(z)$ 在 $z=x+i y$ 处可导, 则

$$
\begin{aligned}
f^{\prime}(z) & =\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=\frac{\partial u}{\partial x}-i \frac{\partial u}{\partial y} \\
& =\frac{\partial v}{\partial y}-i \frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}+i \frac{\partial v}{\partial x}
\end{aligned}

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