# 柯西积分公式

\begin{array}{ll}&\oint_{C}\frac{f(z)}{z-z_0}dz-\oint_{|z-z_0|=R}\frac{f(z)}{z-z_0}dz=0\\ \Rightarrow&\oint_{C}\frac{f(z)}{z-z_0}dz=\oint_{|z-z_0|=R}\frac{f(z)}{z-z_0}dz\end{array}

\begin{align*}\int_{|z-z_0|=R}\frac{f(z)}{z-z_0}dz&=\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)+f(z_0)}{z-z_0}dz\\ &=\oint_{|z-z_0|=R}\frac{f(z_0)}{z-z_0}dz+\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz\\ &=f(z_0)\oint_{|z-z_0|=R}\frac{1}{z-z_0}dz+\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz\\ &=f(z_0)\oint_0^{2\pi}\frac{iRe^{i\theta}d\theta}{Ee^{i\theta}}+\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz\\ &=2\pi if(z_0)+\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz\end{align*}

\begin{align*}\left|\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz\right|&\le\oint_{|z-z_0|=R}\frac{|f(z)-f(z_0)|}{|z-z_0|}|dz|\\ &<\frac{\epsilon}{R}\int_0^{2\pi}Rd\theta=2\pi\epsilon\end{align*}

$\oint_{|z-z_0|=R}\frac{f(z)-f(z_0)}{z-z_0}dz$

\begin{align*}\oint_{|z|=2}\frac{e^z}{z-1}dz=2\pi if(1)=2\pi i e\end{align*}

\begin{align*}\oint_{|z|=2}\frac{zdz}{(9-z^2)(z+i)}&=\oint_{|z|=2}\frac{z/(9-z^2)dz}{z+i}\\ &=2\pi i\frac{z}{(9-z^2)}\Big|_{z=-i}\\ &=2\pi i\frac{-i}{9+1}=\frac{\pi}{5}\end{align*}

3，若分母有多个零点在围线内部，就要对每一个零点应用柯西积分公式。

\begin{align*}\oint_C\frac{dz}{(z^2+1)(z^2+4)}&=\oint_C\frac{dz}{(z+i)(z-i)(z^2+4)}\\ &=2\pi i\frac{1}{(z-i)(z^2+4)}\Big|_{z=-i}+2\pi i\frac{1}{(z+i)(z^2+4)}\Big|_{z=i}\\ &=\frac{2\pi i}{-6i}+\frac{2\pi i}{6i}=0\end{align*}

（2）$$|z|=3$$，分母有四个零点在围线内，$$z=\pm i, z=\pm 2i$$，所以

\begin{align*}\oint_C\frac{dz}{(z^2+1)(z^2+4)}&=\oint_C\frac{dz}{(z+i)(z-i)(z+2i)(z-2i)}\\ &=2\pi i\left(\frac{1}{(z-i)(z^2+4)}\Big|_{z=-i}+\frac{1}{(z+i)(z^2+4)}\Big|_{z=i}+\frac{1}{(z^2+1)(z-2i)}\Big|_{z=-2i}+\frac{dz}{(z^2+1)(z+2i)}\Big|_{z=2i}\right)\\ &=2\pi i\left(\frac{1}{-6i}+\frac{1}{6i}+\frac{1}{-12i}+\frac{1}{12i}\right)=0\end{align*}