记 $z_0=x_0+iy_0$. $u_0=u(x_0,y_0)$, $v_0=v(x_0,y_0)$. 则 $f(z)=u+iv$, $f(z_0)=u_0+iv_0$.
\[
\begin{split}
\frac{f(z)-f(z_0)}{z-z_0}&=\frac{(f(z)-f(z_0))\cdot\overline{z-z_0}}{\|z-z_0\|^2}\\
&=\frac{\bigl((u-u_0)+i(v-v_0)\bigr)\bigl((x-x_0)-i(y-y_0)\bigr)}{(x-x_0)^2+(y-y_0)^2}\\
&=\frac{\bigl[(u-u_0)(x-x_0)+(v-v_0)(y-y_0)\bigr]+i\bigl[(v-v_0)(x-x_0)-(u-u_0)(y-y_0)\bigr]}{(x-x_0)^2+(y-y_0)^2}
\end{split}
\]
记 $\rho=\|z-z_0\|=\sqrt{(x-x_0)^2+(y-y_0)^2}$, $\cos\theta=\frac{|x-x_0|}{\rho}$, $\sin\theta=\frac{|y-y_0|}{\rho}$.
因此极限 $\lim\limits_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}$ 存在当且仅当
\[
\lim_{\rho\rightarrow 0}\frac{(u-u_0)(x-x_0)+(v-v_0)(y-y_0)}{\rho^2}=\lim_{\rho\rightarrow 0}\Bigl(\frac{u-u_0}{x-x_0}\cos^2\theta+\frac{v-v_0}{y-y_0}\sin^2\theta\Bigr)
\]
\[
\lim_{\rho\rightarrow 0}\frac{(v-v_0)(x-x_0)-(u-u_0)(y-y_0)}{\rho^2}=\lim_{\rho\rightarrow 0}\Bigl(\frac{v-v_0}{x-x_0}\cos^2\theta-\frac{u-u_0}{y-y_0}\sin^2\theta\Bigr)
\]
均存在.
假设 $f(z)$ 是 $D$ 中的全纯函数, 即对于每个 $z_0\in D$, $f'(z_0)$ 存在.
由于极限存在与 $\theta$ 的取值无关, 因此选择 $z\rightarrow z_0$ 的特殊的趋近方式, 比如, 令 $z=x+iy_0$, 则 $\theta=0$, 可推出
\[
f'(z_0)=\lim_{x\rightarrow x_0}\frac{u-u_0}{x-x_0}+i\lim_{x\rightarrow x_0}\frac{v-v_0}{x-x_0}=(u_x+iv_x)\biggr|_{(x_0,y_0)}.
\]
另一方面, 令 $z=x_0+iy$, 则 $\theta=\frac{\pi}{2}$, 可得
\[
f'(z_0)=\lim_{y\rightarrow y_0}\frac{v-v_0}{y-y_0}-i\lim_{y\rightarrow y_0}\frac{u-u_0}{y-y_0}=(v_y-iu_y)\biggr|_{(x_0,y_0)}.
\]
因此, 得 $u_x+iv_x=v_y-iu_y$, 即有 $u_x=v_y$, $u_y=-v_x$.
反之, 假设 $f(z)$ 的实部和虚部 $u,v$ 满足 Cauchy-Riemann 方程, 即
\[
\begin{cases}
u_x=v_y,\\
u_y=-v_x.
\end{cases}
\]
则
\[
\lim_{\rho\rightarrow 0}\frac{(u-u_0)(x-x_0)+(v-v_0)(y-y_0)}{\rho^2}=\lim_{\rho\rightarrow 0}\Bigl(\frac{u-u_0}{x-x_0}\cos^2\theta+\frac{v-v_0}{y-y_0}\sin^2\theta\Bigr)=u_x(x_0,y_0)
\]
\[
\lim_{\rho\rightarrow 0}\frac{(v-v_0)(x-x_0)-(u-u_0)(y-y_0)}{\rho^2}=\lim_{\rho\rightarrow 0}\Bigl(\frac{v-v_0}{x-x_0}\cos^2\theta-\frac{u-u_0}{y-y_0}\sin^2\theta\Bigr)=v_x(x_0,y_0)
\]
极限都存在. 从而由上面的分析知 $f'(z_0)$ 存在. 由 $z_0$ 的任意性, 故 $f(z)$ 在 $D$ 中全纯.
此外, 我们还推出了
\[
f'(z)=u_x+iv_x=v_y-iu_y
\]
$\frac{\partial}{\partial x}f=\frac{\partial}{\partial x}(u+iv)=u_x+iv_x$, $\frac{\partial}{\partial y}f=\frac{\partial}{\partial y}(u+iv)=u_y+iv_y$. 因此
\[
\frac{1}{2}(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y})=\frac{1}{2}\bigl(u_x+iv_x-i(u_y+iv_y)\bigr)=\frac{1}{2}(u_x+iv_x-i(-v_x+iu_x))=\frac{1}{2}(2u_x+i2v_x)=u_x+iv_x
\]
因此
\[
\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})=\frac{\partial}{\partial z}
\]
另一方面
\[
\begin{split}
&\frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y})\\
=&\frac{1}{2}\bigl(u_x+iv_x+i(u_y+iv_y)\bigr)\\
=&\frac{1}{2}(u_x-v_y+i(v_x+u_y))\\
=&0
\end{split}
\]
因此,
\[
\frac{\partial f}{\partial\bar{z}}=\frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y})=0
\]
换个角度,
\[
\begin{split}
\frac{f(z)-f(z_0)}{\bar{z}-\bar{z_0}}&=\frac{(f(z)-f(z_0))\cdot(z-z_0)}{\|z-z_0\|^2}\\
&=\frac{\bigl((u-u_0)+i(v-v_0)\bigr)\bigl((x-x_0)+i(y-y_0)\bigr)}{(x-x_0)^2+(y-y_0)^2}\\
&=\frac{\bigl[(u-u_0)(x-x_0)-(v-v_0)(y-y_0)\bigr]+i\bigl[(v-v_0)(x-x_0)+(u-u_0)(y-y_0)\bigr]}{(x-x_0)^2+(y-y_0)^2}
\end{split}
\]
