将 $\left\{\begin{array}{l}x=\frac{z+z^*}{2} \\ y=-i \frac{z-z^*}{2}\end{array}\right.$ 表示成 $f\left(z, z^*\right)$（其实任何二元函数都可以这样表示），下面我们来证明对 $z, z^*$ 求导可写成：
$$
\left\{\begin{array}{c}
\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right) \\
\frac{\partial}{\partial z^*}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i \frac{\partial}{\partial y}\right)
\end{array}\right.
$$
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