Laurent級數(shù)與留數(shù)定理

2022-01-01 16:02 作者:子瞻Louis 0人讀過 | 我要投稿

本期專欄的內(nèi)容主要是復(fù)分析的幾個基本公式

本期的定理要用到一個數(shù)學(xué)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 $P%3DP(x%2Cy)%2CQ%3DQ(x%2Cy)$ 是區(qū)域 $z_0$ 上具有一階連續(xù)偏導(dǎo)數(shù)的函數(shù)， $%5Cbar%20D$ 是 $D$ 的邊界正向曲線

本期專欄不給出證明（懶狗本狗）

復(fù)積分中的Cauchy定理

引：

設(shè) $z%3Dx%2Biy%2Cx%2Cy%5Cin%5Cmathbb%20R$ ， $%5COmega%20$ 內(nèi)的全純函數(shù) $f(z)$ 的實部和虛部分別為

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

則Cauchy-Riemann方程（C-R方程）成立：

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

?設(shè) $z_0%3Dx_0%2Biy_0%5Cin%5COmega$ ，因 $f(z)$ 在 $%5COmega%20$ 內(nèi)全純，故存在兩個復(fù)數(shù) $A%2CB$ 使得

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

其中 $z%5Crightarrow%20z_0$ 時， $%5Crho(z)%5Crightarrow0$ ，通過對上式取 $x%2Cy$ 的偏導(dǎo)，可得

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

又根據(jù)

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

$%5Cbar%20z$ 表示 $z$ 的共軛復(fù)數(shù)，由此可得

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

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

觀察紫色部分，發(fā)現(xiàn) $z%5Crightarrow%20z_0$ 時極限不存在：

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

因此，若要讓 $f(z)$ 在 $z_0$ 處解析，必須有

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

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

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

Cauchy定理：

若 $f(z)$ 在區(qū)域D內(nèi)全純，則對于簡單閉曲線 $C%5Csubset%20D$ （連續(xù)而不自交并且能定義長度的閉合曲線），有

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

?設(shè) $f(z)%3Du%2Biv$ ，則

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

利用Green公式，可得

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

再根據(jù)C-R方程，藍色部分恒為0，定理得證?

根據(jù)Cauchy定理，可得一個經(jīng)典的結(jié)論：

$f(z)$ 為D內(nèi)的解析函數(shù)，對于D內(nèi)的兩條起點與終點都重合的簡單閉曲線C1,C2，有

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

這是因為沿C2的負方向剛好與C1共同構(gòu)成一條簡單閉曲線

此推論也可簡訴為復(fù)積分的結(jié)果與積分路徑無關(guān)

Cauchy積分公式

（沒錯本期專欄就是Cauchy主場）

同樣先給個引理：

a是簡單閉曲線C內(nèi)部的一點，則

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

?根據(jù)Cauchy定理，該積分與C的形狀無關(guān)，不妨設(shè)C是以 $z_0$ 為圓心r為半徑的圓周，則

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

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

我們熟知當(dāng)n≠0時，上面的積分等于0，n=0時，上訴積分為2πi，引理得證?

Cauchy積分公式

若 $f(z)$ 在簡單閉曲線C的內(nèi)部全純，a是C內(nèi)部的一點，則

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

?以a點為圓心作一半徑為 $%5Cvarepsilon$ 的圓周 $%5Cgamma$ ，連接 $%5Cvarepsilon$ 與C上的兩點AB

沿 $C%2BBA%2B%5Cgamma%5E-%2BAB$ 積分， $f(z)$ 在被圓環(huán)區(qū)域內(nèi)全純，根據(jù)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與BA的積分剛好可以抵消，于是

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

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

半徑趨于0時右側(cè)積分基本上等于 $2%5Cpi%20if(a)$ ，這里就簡單驗證一下：設(shè)

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

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

推論：Cauchy高階導(dǎo)公式

設(shè) $f(z)$ 在區(qū)域D內(nèi)全純，C是D內(nèi)的一條簡單閉曲線，根據(jù)Cauchy可得它有以下積分表示：

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

對它取n階導(dǎo)數(shù)，可得

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

這就是Cauchy高階導(dǎo)公式了

Laurent級數(shù)

在小學(xué)二年級就已經(jīng)學(xué)過的微積分中，我們知道了滿足一定條件的一個函數(shù)存在以下Taylor級數(shù)展開：

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

but它也有它的局限性，比如 $e%5E%7B1%2Fx%7D$ 在零處就不能展開為Taylor級數(shù)了，不過我們可以通過換元來解決這個問題：

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

這樣雖然展開中出現(xiàn)了負冪次，但除x=0外它右側(cè)級數(shù)都是收斂的

此類展開的推廣就是Laurent級數(shù)（洛朗級數(shù)）展開：

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

Laurent級數(shù)一般被分為兩部分：

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

藍色部分是它的主要部分，紫色部分則是次要部分，這兩個東西本身都是冪級數(shù)，所以它們有其對應(yīng)的收斂域，這里假設(shè)它的收斂域為：

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

這個是復(fù)平面上的一個圓環(huán)，這也告訴了我們半純函數(shù)在一個圓環(huán)內(nèi)全純，那么它就可以在這個圓環(huán)內(nèi)展開為Laurent級數(shù)，

設(shè)一圍道： $%5CGamma%3DC%2BBA%2B%5Cgamma%5E-%2BAB$

根據(jù)Cauchy積分公式，可得

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

同樣AB和BA的積分可以相抵，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是圓環(huán)內(nèi)部的一點，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，積分與和號可以交換順序是由于上訴級數(shù)收斂且圍道積分總是存在

用同樣的方法，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

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

留數(shù)定理

對一個區(qū)域D內(nèi)的僅有 $z_0$ 一個極點（即不解析點）的半純函數(shù) $f(z)$ ，對其Laurent展開后在圍道 $C%5Csubset%20D$ 上積分

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

前面已經(jīng)得知了右側(cè)圍道積分僅當(dāng)n=-1時不為零，故

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

其中， $a_%7B-1%7D$ 就被稱為留數(shù)或殘數(shù)（Residue）????

記? $z_0$ ?處的留數(shù)為 $%5Cmathop%7B%5Ctext%7BRes%7D%7D_%7Bz%3Dz_0%7Df(z)$ ，由此我們引出以下定理：

若 $f(z)$ 是區(qū)域D內(nèi)的亞純函數(shù)，令A(yù)為它在D內(nèi)的極點點集，C是D的邊界曲線，則

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

?設(shè) $%5Cgamma$ 是以極點a為圓心半徑為r的圓周， $%5CGamma$ 是所以這樣的 $%5Cgamma$ 的集合

運用柯西定理，可得

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

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

將右側(cè)每個積分在極點處展開為Laurent級數(shù)，再令 $r%5Crightarrow0$ 即可得證?

運用留數(shù)定理我們可以通過計算一些亞純函數(shù)的留數(shù)來輕易計算圍道積分

這些亞純函數(shù)需要滿足的條件就是Laurent展開中只有有限項負冪次，假設(shè)

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

m為處的極點階數(shù)（即最高負冪次），在左右兩邊乘以 $(z-z_0)%5E%7Bm%7D$ ，所有項就全部是正冪次了，

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

第m-1項就是它的留數(shù)了，可以通過類似計算Taylor級數(shù)系數(shù)的方法算出第n-1項為

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

若最高負冪次是-1，那么該極點稱為單極點，單極點處的留數(shù)為

$%5Cmathop%7B%5Ctext%7BRes%7D%7D_%7Bz%3Dz_0%7Df(z)%3D%5Clim_%7Bz%5Cto%20z_0%7D(z-z_0)f(z)$

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Laurent級數(shù)與留數(shù)定理

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Cauchy積分公式

Laurent級數(shù)

留數(shù)定理