# 解析函数的高阶导数

1，定理（解析函数的高阶导数）：若 $$f(z)$$ 在简单闭曲线 $$C$$ 内解析，$$z_0$$ 为 $$C$$ 内部一点，则

$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz, n=1,2,\cdots$

（1）由柯西积分公式

$f(z_0)\frac{1}{2\pi}=\oint_C\frac{f(z)}{z-z_0}dz,\quad f(z_0+h)=\frac{1}{2\pi}\oint_C\frac{f(z)}{z-(z_0+h)}dz$

\begin{align*}\frac{f(z_0+h)-f(z_0)}{h}&=\frac{1}{2\pi i h}\oint_C\frac{f(z)}{z-(z_0+h)}dz-\frac{1}{2\pi i h}\oint_C\frac{f(z)}{z-z_0}dz\\ &=\frac{1}{2\pi i h}\oint_C\frac{hf(z)}{(z-z_0)(z-z_0-h)}dz\\ &=\frac{1}{2\pi i }\oint_C\frac{f(z)}{(z-z_0)(z-z_0-h)}dz\end{align*}

\begin{align*}\frac{f(z_0+h)-f(z_0)}{h}&-\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz\\ &=\frac{1}{2\pi i }\oint_C\frac{f(z)}{(z-z_0)(z-z_0-h)}dz-\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz\\ &=\frac{1}{2\pi i}\oint_C\frac{(z-z_0)-(z-z_0-h)}{(z-z_0)^2(z-z_0-h)}f(z)dz\\ &=\frac{1}{2\pi i}\oint_C\frac{hf(z)}{(z-z_0)^2(z-z_0-h)}dz\end{align*}

\begin{align*}\left|\frac{f(z_0+h)-f(z_0)}{h}-\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz\right|&=\left|\frac{1}{2\pi i}\oint_C\frac{hf(z)}{(z-z_0)^2(z-z_0-h)}dz\right|\\ &\le \frac{|h|}{2\pi}\cdot\frac{ML}{d^3}\end{align*}

$\frac{f(z_0+h)-f(z_0)}{h}-\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz\to 0$

$f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz$

（2）由前一步

$f'(z_0+h)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0-h)^2}, \quad f'(z_0)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz$

$f^{\prime\prime}(z_0)=\frac{2}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^3}dz$

（3）由数学归纳法，完全类似于第一步，可以证明

$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz, n=1,2,\cdots$

\begin{align*}\oint_{|z|=1}\frac{\cosh z}{z^4}dz&=\frac{2\pi i}{3!}(\cosh z)^{(3)}\Big|_{z=0}\\ &=\frac{2\pi i }{6}\cdot 0=0\end{align*}