# 留数

## 例子

${\displaystyle \oint _{C}{e^{z} \over z^{5}}\,dz}$

${\displaystyle \oint _{C}{1 \over z^{5}}\left(1+z+{z^{2} \over 2!}+{z^{3} \over 3!}+{z^{4} \over 4!}+{z^{5} \over 5!}+{z^{6} \over 6!}+\cdots \right)\,dz.}$

${\displaystyle \oint _{C}\left({1 \over z^{5}}+{z \over z^{5}}+{z^{2} \over 2!\;z^{5}}+{z^{3} \over 3!\;z^{5}}+{z^{4} \over 4!\;z^{5}}+{z^{5} \over 5!\;z^{5}}+{z^{6} \over 6!\;z^{5}}+\cdots \right)\,dz}$
${\displaystyle =\oint _{C}\left({1 \over \;z^{5}}+{1 \over \;z^{4}}+{1 \over 2!\;z^{3}}+{1 \over 3!\;z^{2}}+{1 \over 4!\;z}+{1 \over \;5!}+{z \over 6!}+\cdots \right)\,dz.}$

${\displaystyle \oint _{C}{1 \over z^{a}}\,dz=0,\quad a\in \mathbb {Z} ,\quad a\neq 1.}$

${\displaystyle \oint _{C}{1 \over 4!\;z}\,dz={1 \over 4!}\oint _{C}{1 \over z}\,dz={1 \over 4!}(2\pi i)={\pi i \over 12}.}$

1/4!就是ez/z5z = 0的留数，记为：

${\displaystyle \mathrm {Res} _{0}{e^{z} \over z^{5}},}$${\displaystyle \mathrm {Res} _{z=0}{e^{z} \over z^{5}},}$${\displaystyle \mathrm {Res} (f,0).}$

## 留数的计算

${\displaystyle \operatorname {Res} (f,c)={1 \over 2\pi i}\oint _{\gamma }f(z)\,dz}$

### 可去奇点

#### 一阶极点

${\displaystyle \operatorname {Res} (f,c)=\lim _{z\to c}(z-c)f(z).}$

ghc的一个邻域内是全纯函数，h(c) = 0而g(c) ≠ 0，那么函数f(z)=g(z)/h(z)在极点c的留数为：

${\displaystyle \operatorname {Res} (f,c)={\frac {g(c)}{h'(c)}}.}$

#### 较高阶极点的极限公式

${\displaystyle \mathrm {Res} (f,c)={\frac {1}{(n-1)!}}\lim _{z\to c}{\frac {d^{n-1}}{dz^{n-1}}}\left((z-c)^{n}f(z)\right).}$

### 无穷远点的留数

${\displaystyle \mathrm {Res} (f(z),\infty )=-\mathrm {Res} \left({\frac {1}{z^{2}}}f\left({\frac {1}{z}}\right),0\right)}$.

${\displaystyle \lim _{|z|\to \infty }f(z)=0}$,

${\displaystyle \mathrm {Res} (f,\infty )=-\lim _{|z|\to \infty }z\cdot f(z)}$.

${\displaystyle \lim _{|z|\to \infty }f(z)=c\neq 0}$,

${\displaystyle \mathrm {Res} (f,\infty )=-\lim _{|z|\to \infty }z^{2}\cdot f'(z)}$.

### 级数方法

1. 第一个例子，计算以下函数在奇点的留数：

${\displaystyle f(z)={\sin {z} \over z^{2}-z}}$

${\displaystyle f(z)={\sin {z} \over z(z-1)}}$

${\displaystyle g(z)=g(a)+g'(a)(z-a)+{g''(a)(z-a)^{2} \over 2!}+{g'''(a)(z-a)^{3} \over 3!}+\cdots }$

${\displaystyle \sin {z}=\sin {1}+\cos {1}(z-1)+{-\sin {1}(z-1)^{2} \over 2!}+{-\cos {1}(z-1)^{3} \over 3!}+\cdots .}$

${\displaystyle {\frac {1}{z}}={\frac {1}{(z-1)+1}}=1-(z-1)+(z-1)^{2}-(z-1)^{3}+\cdots .}$

${\displaystyle {\frac {\sin {z}}{z(z-1)}}={\sin {1} \over z-1}+(\cos {1}-\sin 1)+(z-1)\left(-{\frac {\sin {1}}{2!}}-\cos 1+\sin 1\right)+\cdots .}$

2. 接下来的例子展示了运用级数展开来求留数，拉格朗日反演定理在这里发挥了重要作用。令

${\displaystyle u(z):=\sum _{k\geq 1}u_{k}z^{k}}$

${\displaystyle v(z):=\sum _{k\geq 1}v_{k}z^{k}}$

## 参考文献

• Ahlfors, Lars. Complex Analysis. McGraw Hill. 1979.
• Marsden & Hoffman, Basic complex analysis (Freeman, 1999).