Laurent级数与留数定理

2022-01-01 16:02 作者:子瞻Louis 0人读过 | 我要投稿

本期专栏的内容主要是复分析的几个基本公式

本期的定理要用到一个数学分析中的公式，即Green公式：

$%5Ciint_D%5Cleft(%5Cfrac%7B%5Cpartial%20P%7D%7B%5Cpartial%20x%7D-%5Cfrac%7B%5Cpartial%20Q%7D%7B%5Cpartial%20y%7D%5Cright)%5Cmathrm%20dx%5Cmathrm%20dy%3D%5Coint_%5Cbar%20D%20P%5Cmathrm%20dy%2BQ%5Cmathrm%20dx$

其中 $P%3DP(x%2Cy)%2CQ%3DQ(x%2Cy)$ 是区域 $z_0$ 上具有一阶连续偏导数的函数， $%5Cbar%20D$ 是 $D$ 的边界正向曲线

本期专栏不给出证明（懒狗本狗）

复积分中的Cauchy定理

引：

设 $z%3Dx%2Biy%2Cx%2Cy%5Cin%5Cmathbb%20R$ ， $%5COmega%20$ 内的全纯函数 $f(z)$ 的实部和虚部分别为

$f(x%2Biy)%3Du(x%2Cy)%2Biv(x%2Cy)$

则Cauchy-Riemann方程（C-R方程）成立：

$%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%2C%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%3D-%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D$

◀设 $z_0%3Dx_0%2Biy_0%5Cin%5COmega$ ，因 $f(z)$ 在 $%5COmega%20$ 内全纯，故存在两个复数 $A%2CB$ 使得

$f(z)-f(z_0)%3DA(x-x_0)%2BB(y-y_0)%2B%5Crho(z)(z-z_0)$

其中 $z%5Crightarrow%20z_0$ 时， $%5Crho(z)%5Crightarrow0$ ，通过对上式取 $x%2Cy$ 的偏导，可得

$%5Cbegin%7Baligned%7DA%26%3D%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D(z_0)%3D%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(x_0%2Cy_0)%2Bi%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D(x_0%2Cy_0)%5C%5CB%26%3D%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D(z_0)%3D%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D(x_0%2Cy_0)%2Bi%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D(x_0%2Cy_0)%5Cend%7Baligned%7D$

又根据

$z%3Dx%2Biy%2C%5Cbar%20z%3Dx-iy%5CRightarrow%20x%3D%5Cfrac%7Bz%2B%5Cbar%20z%7D2%2Cy%3D%5Cfrac%7Bz-%5Cbar%20z%7D%7B2i%7D$

$%5Cbar%20z$ 表示 $z$ 的共轭复数，由此可得

$%5Cbegin%7Baligned%7Df(z)-f(z_0)%26%3DA%5Cleft(%5Cfrac%7Bz%2B%5Cbar%20z%7D2-%5Cfrac%7Bz_0%2B%5Cbar%7Bz_0%7D%7D2%5Cright)%2BB%5Cleft(%5Cfrac%7Bz-%5Cbar%20z%7D%7B2i%7D-%5Cfrac%7Bz_0-%5Cbar%7Bz_0%7D%7D%7B2i%7D%5Cright)%2B%5Crho(z)(z-z_0)%5C%5C%26%3D%5Cfrac%7BA-iB%7D2(z-z_0)%2B%5Cfrac%7BA%2BiB%7D2(%5Cbar%20z-%5Cbar%7Bz_0%7D)%2B%5Crho(z)(z-z_0)%5Cend%7Baligned%7D$

$%5CRightarrow%5Cfrac%7Bf(z)-f(z_0)%7D%7Bz-z_0%7D%3D%5Cfrac%7BA-iB%7D2%2B%5Crho(z)%2B%0A%5Ccolor%7Bpurple%7D%7B%5Cfrac%7BA%2BiB%7D2%5Cfrac%7B%5Coverline%7Bz-z_0%7D%7D%7Bz-z_0%7D%7D$

观察紫色部分，发现 $z%5Crightarrow%20z_0$ 时极限不存在：

$%5Clim_%7Bz%5Cto%20z_0%5C%5C%5CIm%7B(z-z_0)%7D%3D0%7D%5Cfrac%7B%5Coverline%7Bz-z_0%7D%7D%7Bz-z_0%7D%3D1%E2%89%A0%5Clim_%7Bz%5Cto%20z_0%5C%5C%5CRe%7B(z-z_0)%7D%3D0%7D%5Cfrac%7B%5Coverline%7Bz-z_0%7D%7D%7Bz-z_0%7D%3D-1$

因此，若要让 $f(z)$ 在 $z_0$ 处解析，必须有

$%5Cfrac%7BA%2BiB%7D2%3D0$

$%5CRightarrow%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D(x_0%2Cy_0)-%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D(x_0%2Cy_0)%2Bi%5Cleft(%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D(x_0%2Cy_0)%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D(x_0%2Cy_0)%5Cright)%3D0$

$%5CRightarrow%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%2C%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%3D-%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D$ ▶

Cauchy定理：

若 $f(z)$ 在区域D内全纯，则对于简单闭曲线 $C%5Csubset%20D$ （连续而不自交并且能定义长度的闭合曲线），有

$%5Coint_Cf(z)%5Cmathrm%20dz%3D0$

◀设 $f(z)%3Du%2Biv$ ，则

$%5Cbegin%7Baligned%7D%5Coint_Cf(z)%5Cmathrm%20dz%26%3D%5Coint_C(u%2Biv)%5Cmathrm%20d(x%2Biy)%5C%5C%26%3D%5Coint_Cu%5Cmathrm%20dx-v%5Cmathrm%20dy%2Bi%5Coint_Cu%5Cmathrm%20dy%2Bv%5Cmathrm%20dx%5Cend%7Baligned%7D$

利用Green公式，可得

$%5Cbegin%7Baligned%7D%5Coint_Cf(z)%5Cmathrm%20dz%26%3D%5Coint_Cu%5Cmathrm%20dx-v%5Cmathrm%20dy%2Bi%5Coint_Cu%5Cmathrm%20dy%2Bv%5Cmathrm%20dx%5C%5C%26%3D%5Ciint_D%5Ccolor%7Bblue%7D%7B%5Cleft(%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D%5Cright)%7D%5Cmathrm%20dx%5Cmathrm%20dy%2Bi%5Ciint_D%5Ccolor%7Bblue%7D%7B%5Cleft(%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D-%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%5Cright)%7D%5Cmathrm%20dx%5Cmathrm%20dy%5Cend%7Baligned%7D$

再根据C-R方程，蓝色部分恒为0，定理得证▶

根据Cauchy定理，可得一个经典的结论：

$f(z)$ 为D内的解析函数，对于D内的两条起点与终点都重合的简单闭曲线C1,C2，有

$%5Cint_%7BC_1%7Df(z)%5Cmathrm%20dz%3D%5Cint_%7BC_2%7Df(z)%5Cmathrm%20dz$

这是因为沿C2的负方向刚好与C1共同构成一条简单闭曲线

此推论也可简诉为复积分的结果与积分路径无关

Cauchy积分公式

（没错本期专栏就是Cauchy主场）

同样先给个引理：

a是简单闭曲线C内部的一点，则

$%5Coint_C%5Cfrac%7B%5Cmathrm%20dz%7D%7B(z-a)%5E%7Bn%2B1%7D%7D%3D%5Cbegin%7Bcases%7D%202%5Cpi%20i%26n%3D0%5C%5C0%20%26%20n%E2%89%A00%2Cn%5Cin%5Cmathbb%20Z%5Cend%7Bcases%7D$

◀根据Cauchy定理，该积分与C的形状无关，不妨设C是以 $z_0$ 为圆心r为半径的圆周，则

$z%3Da%2Bre%5E%7Bit%7D%2Ct%3A0%5Crightarrow2%5Cpi%2C%5Cmathrm%20dz%3Dire%5E%7Bit%7D%5Cmathrm%20dt$

$%5Cbegin%7Baligned%7D%5Coint_C%5Cfrac%7B%5Cmathrm%20dz%7D%7B(z-a)%5E%7Bn%2B1%7D%7D%26%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cfrac%7Bire%5E%7Bit%7D%7D%7Br%5E%7Bn%2B1%7De%5E%7B(n%2B1)it%7D%7D%5Cmathrm%20dt%5C%5C%26%3Di%5Cfrac1%7Br%5En%7D%5Cint_0%5E%7B2%5Cpi%7De%5E%7B-int%7D%5Cmathrm%20dt%5Cend%7Baligned%7D$

我们熟知当n≠0时，上面的积分等于0，n=0时，上诉积分为2πi，引理得证▶

Cauchy积分公式

若 $f(z)$ 在简单闭曲线C的内部全纯，a是C内部的一点，则

$%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%3Df(a)$

◀以a点为圆心作一半径为 $%5Cvarepsilon$ 的圆周 $%5Cgamma$ ，连接 $%5Cvarepsilon$ 与C上的两点AB

沿 $C%2BBA%2B%5Cgamma%5E-%2BAB$ 积分， $f(z)$ 在被圆环区域内全纯，根据Cauchy定理，有

$%5Cint_C%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%2B%5Cint_%7BBA%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%2B%5Cint_%7B%5Cgamma%5E-%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%2B%5Cint_%7BAB%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%3D0$

注意到沿AB与BA的积分刚好可以抵消，于是

$%5Coint_C%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%2B%5Coint_%7B%5Cgamma%5E-%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%3D0$

$%5CRightarrow$ $%5Coint_C%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz%3D%5Coint_%7B%5Cgamma%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz$

半径趋于0时右侧积分基本上等于 $2%5Cpi%20if(a)$ ，这里就简单验证一下：设

$z%3Da%2B%5Cvarepsilon%20e%5E%7Bit%7D%2Ct%3A0%5Crightarrow2%5Cpi%2C%5Cmathrm%20dz%3Di%5Cvarepsilon%20e%5E%7Bit%7D%5Cmathrm%20dt$

$%5Cbegin%7Baligned%7D%5CRightarrow%5Cleft%7C%5Coint_%7B%5Cgamma%7D%5Cfrac%7Bf(z)%7D%7Bz-a%7D%5Cmathrm%20dz-%5Coint_%7B%5Cgamma%7D%5Cfrac%7Bf(a)%7D%7Bz-a%7D%5Cmathrm%20dz%5Cright%7C%26%3D%5Cleft%7C%5Cint_0%5E%7B2%5Cpi%7D%5Cfrac%7Bf(a%2B%5Cvarepsilon%20e%5E%7Bit%7D)-f(a)%7D%7B%5Cvarepsilon%20e%5E%7Bit%7D%7Di%5Cvarepsilon%20e%5E%7Bit%7D%5Cmathrm%20dt%5Cright%7C%5C%5C%26%3D%5Cleft%7C%20i%5Cint_0%5E%7B2%5Cpi%7Df(a%2B%5Cvarepsilon%20e%5E%7Bit%7D)-f(a)%5Cmathrm%20dt%5Cright%7C%5C%5C%26%5Cle%5Cint_0%5E%7B2%5Cpi%7D%5Cvert%20f(a%2B%5Cvarepsilon%20e%5E%7Bit%7D)-f(a)%5Cvert%5Cmathrm%20dt%5C%5C%26%5Cle2%5Cpi%5Cmax_%7B0%5Cle%20%5Cxi%5Cle2%5Cpi%7D%5Cvert%20f(a%2B%5Cvarepsilon%20e%5E%7Bi%5Cxi%7D)-f(a)%5Cvert%5Crightarrow0%5Cend%7Baligned%7D$

▶

推论：Cauchy高阶导公式

设 $f(z)$ 在区域D内全纯，C是D内的一条简单闭曲线，根据Cauchy可得它有以下积分表示：

$f(z)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(w)%7D%7Bw-z%7D%5Cmathrm%20dw$

对它取n阶导数，可得

$%5Cfrac%7B%5Cmathrm%20d%5En%7D%7B%5Cmathrm%20dz%5En%7Df(z)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_Cf(w)%5Cfrac%7B%5Cpartial%5En%7D%7B%5Cpartial%20z%5En%7D%5Cfrac1%7Bw-z%7D%5Cmathrm%20dw%3D%5Cfrac%7Bn!%7D%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(w)%7D%7B(w-z)%5E%7Bn%2B1%7D%7D%5Cmathrm%20dw$

这就是Cauchy高阶导公式了

Laurent级数

在小学二年级就已经学过的微积分中，我们知道了满足一定条件的一个函数存在以下Taylor级数展开：

$f(x)%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7Bf%5E%7B(n)%7D(x_0)%7D%7Bn!%7D(x-x_0)%5En$

but它也有它的局限性，比如 $e%5E%7B1%2Fx%7D$ 在零处就不能展开为Taylor级数了，不过我们可以通过换元来解决这个问题：

$e%5E%20t%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7Bt%5En%7D%7Bn!%7D%5Cstackrel%7Bt%3D1%2Fx%7D%7B%5Clongrightarrow%7De%5E%7B1%2Fx%7D%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7Bx%5E%7B-n%7D%7D%7Bn!%7D$

这样虽然展开中出现了负幂次，但除x=0外它右侧级数都是收敛的

此类展开的推广就是Laurent级数（洛朗级数）展开：

$f(z)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20a_n(z-z_0)%5En$

Laurent级数一般被分为两部分：

$f(z)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20a_n(z-z_0)%5En%3D%5Ccolor%7Bblue%7D%7B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20a_%7B-n%7D(z-z_0)%5E%7B-n%7D%7D%2B%5Ccolor%7Bpurple%7D%7B%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20a_%7Bn%7D(z-z_0)%5E%7Bn%7D%7D$

蓝色部分是它的主要部分，紫色部分则是次要部分，这两个东西本身都是幂级数，所以它们有其对应的收敛域，这里假设它的收敛域为：

$%5Cleft%7C%5Cfrac1%7Bz-z_0%7D%5Cright%7C%5Cle%20r%2C%5Cleft%7Cz-z_0%5Cright%7C%5Cle%20R%5CRightarrow%5Cfrac1r%5Cle%5Cvert%20z-z_0%5Cvert%5Cle%20R$

这个是复平面上的一个圆环，这也告诉了我们半纯函数在一个圆环内全纯，那么它就可以在这个圆环内展开为Laurent级数，

设一围道： $%5CGamma%3DC%2BBA%2B%5Cgamma%5E-%2BAB$

根据Cauchy积分公式，可得

$f(z)%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%7B%5CGamma%7D%5Cfrac%7Bf(s)%7D%7Bs%20-z%7D%5Cmathrm%20ds$

同样AB和BA的积分可以相抵，故

$f(z)%3D%5Cint_%7BC%2B%5Cgamma%5E-%7D%5Cfrac%7Bf(s)%7D%7Bs-z%7D%5Cmathrm%20ds%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(s)%7D%7Bs-z%7D%5Cmathrm%20ds-%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%5Cgamma%5Cfrac%7Bf(s)%7D%7Bs-z%7D%5Cmathrm%20ds$

z是圆环内部的一点，因此在C上， $%7Cz-z_0%7C%EF%BC%9C%7Cs-z_0%7C%5CRightarrow%20%5Cleft%7C%5Cfrac%7Bz-z_0%7D%7Bs-z_0%7D%5Cright%7C%EF%BC%9C1$

$%5Cbegin%7Baligned%7D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(s)%7D%7Bs-z%7D%5Cmathrm%20ds%26%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(s)%7D%7Bs-z_0-(z-z_0)%7D%5Cmathrm%20ds%5C%5C%26%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(s)%7D%7Bs-z_0%7D%5Ccdot%5Cfrac1%7B1-%5Cfrac%7Bz-z_0%7D%7Bs-z_0%7D%7D%5Cmathrm%20ds%5C%5C%26%3D%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%5Cfrac%7Bf(s)%7D%7Bs-z_0%7D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cleft(%5Cfrac%7Bz-z_0%7D%7Bs-z_0%7D%5Cright)%5En%5Cmathrm%20ds%5C%5C%26%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cleft%5C%7B%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_C%20%5Cfrac%7Bf(s)%7D%7B(s-z_0)%5E%7Bn%2B1%7D%7D%5Cmathrm%20ds%5Cright%5C%7D(z-z_0)%5En%5Cend%7Baligned%7D$

其中，积分与和号可以交换顺序是由于上诉级数收敛且围道积分总是存在

用同样的方法，可以得到 $%7Cs-z_0%7C%EF%BC%9C%7Cz-z_0%7C%5CRightarrow%20%5Cleft%7C%5Cfrac%7Bs-z_0%7D%7Bz-z_0%7D%5Cright%7C%EF%BC%9C1$

$-%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%5Cgamma%5Cfrac%7Bf(s)%7D%7Bs-z%7D%5Cmathrm%20ds%3D%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cleft%5C%7B%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%5Cgamma%5Cfrac%7Bf(s)%7D%7B(s-z_0)%5E%7B-n%2B1%7D%7D%5Cmathrm%20ds%5Cright%5C%7D(z-z_0)%5E%7B-n%7D$

代回 $f(z)$ 中，可得

$f(z)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%5Ccolor%7Bred%7D%7B%5Cleft%5C%7B%5Cfrac1%7B2%5Cpi%20i%7D%5Coint_%7B%5CGamma%7D%20%5Cfrac%7Bf(s)%7D%7B(s-z_0)%5E%7Bn%2B1%7D%7D%5Cmathrm%20ds%5Cright%5C%7D%7D(z-z_0)%5En$

留数定理

对一个区域D内的仅有 $z_0$ 一个极点（即不解析点）的半纯函数 $f(z)$ ，对其Laurent展开后在围道 $C%5Csubset%20D$ 上积分

$%5Coint_Cf(z)%5Cmathrm%20dz%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%5Cinfty%20a_n%5Coint_C(z-z_0)%5En%5Cmathrm%20dz$

前面已经得知了右侧围道积分仅当n=-1时不为零，故

$%5Coint_Cf(z)%5Cmathrm%20dz%3D2%5Cpi%20ia_%7B-1%7D$

其中， $a_%7B-1%7D$ 就被称为留数或残数（Residue）

记 $z_0$ 处的留数为 $%5Cmathop%7B%5Ctext%7BRes%7D%7D_%7Bz%3Dz_0%7Df(z)$ ，由此我们引出以下定理：

若 $f(z)$ 是区域D内的亚纯函数，令A为它在D内的极点点集，C是D的边界曲线，则

$%5Coint_Cf(z)%5Cmathrm%20dz%3D%5Csum_%7Ba%5Cin%20A%7D2%5Cpi%20i%5Cmathop%7B%5Cmathrm%7BRes%7D%7D_%7Bz%3Da%7Df(z)$

◀设 $%5Cgamma$ 是以极点a为圆心半径为r的圆周， $%5CGamma$ 是所以这样的 $%5Cgamma$ 的集合

运用柯西定理，可得

$%5Coint_%7BC-%5CGamma%7Df(z)%5Cmathrm%20dz%3D0$

$%5CRightarrow%5Coint_Cf(z)%5Cmathrm%20dz%3D%5Csum_%7B%5Cgamma%5Cin%5CGamma%7D%5Coint_%5Cgamma%20f(z)%5Cmathrm%20dz$

将右侧每个积分在极点处展开为Laurent级数，再令 $r%5Crightarrow0$ 即可得证▶

运用留数定理我们可以通过计算一些亚纯函数的留数来轻易计算围道积分

这些亚纯函数需要满足的条件就是Laurent展开中只有有限项负幂次，假设

$f(z)%3D%5Csum_%7Bn%3D-m%7D%5E%5Cinfty%20a_%7Bn%7D(z-z_0)%5E%7Bn%7D$

m为处的极点阶数（即最高负幂次），在左右两边乘以 $(z-z_0)%5E%7Bm%7D$ ，所有项就全部是正幂次了，

$(z-z_0)%5Emf(z)%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20a_%7Bn-m%7D(z-z_0)%5E%7Bn%7D$

第m-1项就是它的留数了，可以通过类似计算Taylor级数系数的方法算出第n-1项为

$%5Cmathop%7B%5Ctext%7BRes%7D%7D_%7Bz%3Dz_0%7Df(z)%3D%5Clim_%7Bz%5Cto%20z_0%7D%5Cfrac1%7B(m-1)!%7D%5Cfrac%7B%5Cmathrm%20d%5E%7Bm-1%7D%7D%7B%5Cmathrm%20dz%5E%7Bm-1%7D%7D%5Cleft%5B(z-z_0)%5Emf(z)%5Cright%5D$