设 f ( z ) f\left( z \right) f(z) 在圆环域 R 1 < ∣ z − z 0 ∣ < R 2 R_1 < \left| {z - z_0 } \right| < R_2 R1​<∣z−z0​∣<R2​ 内处处解析，那么 f ( z ) = ∑ n = − ∞ + ∞ c n ( z − z 0 ) n f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {c_n \left( {z - z_0 } \right)^n } f(z)=n=−∞∑+∞​cn​(z−z0​)n 其中 c n = 1 2 π i ∮ C f ( ζ ) ( ζ − z 0 ) n + 1 d ζ ,     ( n = 0 , ± 1 , ± 2 , ⋯ ) c_n = {1 \over {2\pi i}}\oint_C { { {f\left( \zeta \right)} \over {\left( {\zeta - z_0 } \right)^{n + 1} }}d\zeta } ,\,\,\,\left( {n = 0, \pm 1, \pm 2, \cdots } \right) cn​=2πi1​∮C​(ζ−z0​)n+1f(ζ)​dζ,(n=0,±1,±2,⋯)

这个公式称为函数 f ( z ) f\left( z \right) f(z) 在以 z 0 z_0 z0​ 为中心的圆环域： R 1 < ∣ z − z 0 ∣ < R 2 R_1 < \left| {z - z_0 } \right| < R_2 R1​<∣z−z0​∣<R2​ 内的洛朗（Laurent）展开式，称 f ( z ) f\left( z \right) f(z) 在此圆环内的洛朗级数。

• 解析部分：洛朗级数中的正整次幂级数部分；
• 主要部分：洛朗级数中的负整次幂级数部分；

  这种展开形式具有唯一性。

  根据洛朗级数系数 c n c_n cn​ 计算公式，令 n = − 1 n = - 1 n=−1 时，则有 c − 1 = 1 2 π i ∮ C f ( z ) d z c_{ - 1} = {1 \over {2\pi i}}\oint_C {f\left( z \right)dz} c−1​=2πi1​∮C​f(z)dz 所以 ∮ C f ( z ) d z = 2 π i ⋅ c − 1 \oint_C {f\left( z \right)dz} = 2\pi i \cdot c_{ - 1} ∮C​f(z)dz=2πi⋅c−1​

  函数 f ( z ) = 1 ( z − 1 ) ( z − 2 ) f\left( z \right) = {1 \over {\left( {z - 1} \right)\left( {z - 2} \right)}} f(z)=(z−1)(z−2)1​ 在以下三种圆环域处处解析，试把 f ( z ) f\left( z \right) f(z) 在这些区域内展成洛朗级数。

  （1） 0 < ∣ z ∣ < 1 ; 0 < \left| z \right| < 1; 0<∣z∣<1;   （2） 1 < ∣ z ∣ < 2 1 < \left| z \right| < 2 1<∣z∣<2 ；   （3） 2 < ∣ z ∣ < + ∞ 2 < \left| z \right| < + \infty 2<∣z∣<+∞

  求解： 先把 f ( z ) f\left( z \right) f(z) 进行分式分解 f ( z ) = 1 1 − z − 1 2 − z f\left( z \right) = {1 \over {1 - z}} - {1 \over {2 - z}} f(z)=1−z1​−2−z1​

  （1） 在 0 < ∣ z ∣ < 1 0 < \left| z \right| < 1 0<∣z∣<1 内，由于 ∣ z ∣ < 1 \left| z \right| < 1 ∣z∣<1 ，从而 ∣ z / 2 ∣ < 1 \left| {z/2} \right| < 1 ∣z/2∣<1 ，所以 1 1 − z = ∑ n = 0 ∞ z n ,    1 2 − z = 1 2 ∑ n = 0 ∞ ( z 2 ) n {1 \over {1 - z}} = \sum\limits_{n = 0}^\infty {z^n } ,\,\,{1 \over {2 - z}} = {1 \over 2}\sum\limits_{n = 0}^\infty {\left( { {z \over 2}} \right)^n } 1−z1​=n=0∑∞​zn,2−z1​=21​n=0∑∞​(2z​)n 因此 f ( z ) = ∑ n = 0 ∞ z n − 1 2 ∑ n = 0 ∞ ( z 2 ) n = ∑ n = 0 ∞ ( 1 − 1 2 n + 1 ) z n f\left( z \right) = \sum\limits_{n = 0}^\infty {z^n } - {1 \over 2}\sum\limits_{n = 0}^\infty {\left( { {z \over 2}} \right)^n } = \sum\limits_{n = 0}^\infty {\left( {1 - {1 \over {2^{n + 1} }}} \right)z^n } f(z)=n=0∑∞​zn−21​n=0∑∞​(2z​)n=n=0∑∞​(1−2n+11​)zn

  结果中不包含 z z z 负幂级数，因为 f ( z ) f\left( z \right) f(z) 在 z = 0 z = 0 z=0 处解析。

  （2） 在 1 < ∣ z ∣ < 2 1 < \left| z \right| < 2 1<∣z∣<2 内，由于 ∣ z ∣ > 1 \left| z \right| > 1 ∣z∣>1 所以 ∣ 1 z ∣ < 1 \left| { {1 \over z}} \right| < 1 ∣∣​z1​∣∣​<1 ，所以 1 1 − z = − 1 z ⋅ 1 1 − 1 z = − 1 z ∑ n = 0 ∞ z − n {1 \over {1 - z}} = - {1 \over z} \cdot {1 \over {1 - {1 \over z}}} = - {1 \over z}\sum\limits_{n = 0}^\infty {z^{ - n} } 1−z1​=−z1​⋅1−z1​1​=−z1​n=0∑∞​z−n 所以 f ( z ) = − 1 z ∑ n = 0 ∞ z − n − 1 2 ∑ n = 0 ∞ ( z 2 ) n f\left( z \right) = - {1 \over z}\sum\limits_{n = 0}^\infty {z^{ - n} } - {1 \over 2}\sum\limits_{n = 0}^\infty {\left( { {z \over 2}} \right)^n } f(z)=−z1​n=0∑∞​z−n−21​n=0∑∞​(2z​)n

  （3） 当 2 < ∣ z ∣ < ∞ 2 < \left| z \right| < \infty 2<∣z∣<∞ 时，对于 1 2 − z {1 \over {2 - z}} 2−z1​ 也需要进行另外一种形式的展开 1 2 − z = − 1 z ⋅ 1 1 − 2 z = − 1 z ∑ n = 0 ∞ ( z 2 ) − n {1 \over {2 - z}} = - {1 \over z} \cdot {1 \over {1 - {2 \over z}}} = - {1 \over z}\sum\limits_{n = 0}^\infty {\left( { {z \over 2}} \right)^{ - n} } 2−z1​=−z1​⋅1−z2​1​=−z1​n=0∑∞​(2z​)−n 所以 f ( z ) = 1 z ∑ n = 0 ∞ ( z 2 ) − n − 1 z ∑ n = 0 ∞ z − n = ∑ n = 0 ∞ ( 2 n − 1 ) z − ( n + 2 ) f\left( z \right) = {1 \over z}\sum\limits_{n = 0}^\infty {\left( { {z \over 2}} \right)^{ - n} - {1 \over z}\sum\limits_{n = 0}^\infty {z^{ - n} } = \sum\limits_{n = 0}^\infty {\left( {2^n - 1} \right)z^{ - \left( {n + 2} \right)} } } f(z)=z1​n=0∑∞​(2z​)−n−z1​n=0∑∞​z−n=n=0∑∞​(2n−1)z−(n+2)

  求下列个积分值

  （1） ∮ ∣ z ∣ = 3 1 z ( z + 1 ) ( z + 4 ) d z \oint_{\left| z \right| = 3} { {1 \over {z\left( {z + 1} \right)\left( {z + 4} \right)}}dz} ∮∣z∣=3​z(z+1)(z+4)1​dz

  （2） ∮ ∣ z ∣ = 2 z e 1 z 1 − z d z \oint_{\left| z \right| = 2} { { {ze^{ {1 \over z}} } \over {1 - z}}dz} ∮∣z∣=2​1−zzez1​​dz

  求解： 将上面积分内的函数进行洛朗级数展开，获得对应 c − 1 c_{ - 1} c−1​ 系数，那么积分值就等于 2 π i ⋅ c − 1 2\pi i \cdot c_{ - 1} 2πi⋅c−1​ 。

  （1） 函数 f ( z ) = 1 z ( z + 1 ) ( z + 4 ) f\left( z \right) = {1 \over {z\left( {z + 1} \right)\left( {z + 4} \right)}} f(z)=z(z+1)(z+4)1​ 在 1 < ∣ z ∣ < 4 1 < \left| z \right| < 4 1<∣z∣<4 内处处解析， ∣ z ∣ = 3 \left| z \right| = 3 ∣z∣=3 圆环域内，所以可以将其展开洛朗级数 f ( z ) = 1 4 z − 1 3 ( z + 1 ) + 1 12 ( z + 4 ) = 1 4 z − 1 3 z ( 1 + 1 z ) + 1 48 ( 1 + z 4 ) f\left( z \right) = {1 \over {4z}} - {1 \over {3\left( {z + 1} \right)}} + {1 \over {12\left( {z + 4} \right)}} = {1 \over {4z}} - {1 \over {3z\left( {1 + {1 \over z}} \right)}} + {1 \over {48\left( {1 + {z \over 4}} \right)}} f(z)=4z1​−3(z+1)1​+12(z+4)1​=4z1​−3z(1+z1​)1​+48(1+4z​)1​ = 1 4 z − 1 3 z + 1 3 z 3 − ⋯ + 1 48 ( 1 − z 4 + z 2 16 − ⋯ ) = {1 \over {4z}} - {1 \over {3z}} + {1 \over {3z^3 }} - \cdots + {1 \over {48}}\left( {1 - {z \over 4} + { {z^2 } \over {16}} - \cdots } \right) =4z1​−3z1​+3z31​−⋯+481​(1−4z​+16z2​−⋯)由此可见 c − 1 = 1 4 − 1 3 = − 1 12 c_{ - 1} = {1 \over 4} - {1 \over 3} = - {1 \over {12}} c−1​=41​−31​=−121​ ,从而 ∮ C d z z ( z + 1 ) ( z + 4 ) = 2 π i ( − 1 12 ) = − π i 6 \oint_C { { {dz} \over {z\left( {z + 1} \right)\left( {z + 4} \right)}}} = 2\pi i\left( { - {1 \over {12}}} \right) = - { {\pi i} \over 6} ∮C​z(z+1)(z+4)dz​=2πi(−121​)=−6πi​

  （2） 把 f ( z ) = z e 1 z 1 − z f\left( z \right) = { {ze^{ {1 \over z}} } \over {1 - z}} f(z)=1−zzez1​​ zl 1 < ∣ z ∣ < + ∞ 1 < \left| z \right| < + \infty 1<∣z∣<+∞ 圆环内洛朗级数展开 f ( z ) = e 1 z − ( 1 − 1 z ) = − ( 1 + 1 z + 1 z 2 + ⋯ ) ( 1 + 1 z + 1 2 ! z 2 + ⋯ ) f\left( z \right) = { {e^{ {1 \over z}} } \over { - \left( {1 - {1 \over z}} \right)}} = - \left( {1 + {1 \over z} + {1 \over {z^2 }} + \cdots } \right)\left( {1 + {1 \over z} + {1 \over {2!z^2 }} + \cdots } \right) f(z)=−(1−z1​)ez1​​=−(1+z1​+z21​+⋯)(1+z1​+2!z21​+⋯) = − ( 1 + 2 z + 5 2 z 2 + ⋯ ) = - \left( {1 + {2 \over z} + {5 \over {2z^2 }} + \cdots } \right) =−(1+z2​+2z25​+⋯)

  所以 c − 1 = − 2 c_{ - 1} = - 2 c−1​=−2 ，从而 ∮ C z e 1 z 1 − z d z = 2 π i ⋅ c − 1 = − 4 π i \oint_C { { {ze^{ {1 \over z}} } \over {1 - z}}dz} = 2\pi i \cdot c_{ - 1} = - 4\pi i ∮C​1−zzez1​​dz=2πi⋅c−1​=−4πi

  对于离散序列 x [ n ] x\left[ n \right] x[n] 进行z变换，所得到的结果 X ( z ) X\left( z \right) X(z) 实际上就是在其收敛域内的洛朗级数展开。 X ( z ) = ∑ n = − ∞ + ∞ x [ n ] z − n X\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {x\left[ n \right]z^{ - n} } X(z)=n=−∞∑+∞​x[n]z−n

  通过前面洛朗级数展开公式可以获得 z z z 反变换公式。

  将下列函数在指定的圆环域内展开成洛朗级数。

( 1 )        1 ( z 2 + 1 ) ( z − 2 ) ,     1 < ∣ z ∣ < 2 \left( 1 \right)\,\,\,\,\,\,{1 \over {\left( {z^2 + 1} \right)\left( {z - 2} \right)}},\,\,\,1 < \left| z \right| < 2 (1)(z2+1)(z−2)1​,1<∣z∣<2 ( 2 )      1 z ( 1 − z ) 2 ,     0 < ∣ z ∣ < 2 ;     0 < ∣ z − 1 ∣ < 1 ; \left( 2 \right)\,\,\,\,{1 \over {z\left( {1 - z} \right)^2 }},\,\,\,0 < \left| z \right| < 2;\,\,\,0 < \left| {z - 1} \right| < 1; (2)z(1−z)21​,0<∣z∣<2;0<∣z−1∣<1; ( 3 )      1 ( z − 1 ) ( z − 2 ) ,    0 < ∣ z − 1 ∣ < 1 ;     1 < ∣ z − 2 ∣ < + ∞ \left( 3 \right)\,\,\,\,{1 \over {\left( {z - 1} \right)\left( {z - 2} \right)}},\,\,0 < \left| {z - 1} \right| < 1;\,\,\,1 < \left| {z - 2} \right| < + \infty (3)(z−1)(z−2)1​,0<∣z−1∣<1;1<∣z−2∣<+∞ ( 4 )      e 1 1 − z ,     1 < ∣ z ∣ < + ∞ \left( 4 \right)\,\,\,\,e^{ {1 \over {1 - z}}} ,\,\,\,1 < \left| z \right| < + \infty (4)e1−z1​,1<∣z∣<+∞ ( 5 )     1 z 2 ( z − i ) ,     ∣ z − i ∣ < R \left( 5 \right)\,\,\,{1 \over {z^2 \left( {z - i} \right)}},\,\,\,\left| {z - i} \right| < R (5)z2(z−i)1​,∣z−i∣<R ( 6 )      sin ⁡ 1 1 − z ,      ∣ z − 1 ∣ > 0 \left( 6 \right)\,\,\,\,\sin {1 \over {1 - z}},\,\,\,\,\left| {z - 1} \right| > 0 (6)sin1−z1​,∣z−1∣>0 ( 7 )      ( z − 1 ) ( z − 2 ) ( z − 3 ) ( z − 4 ) ,     3 < ∣ z ∣ < 4 ;    4 < ∣ z ∣ < + ∞ \left( 7 \right)\,\,\,\,{ {\left( {z - 1} \right)\left( {z - 2} \right)} \over {\left( {z - 3} \right)\left( {z - 4} \right)}},\,\,\,3 < \left| z \right| < 4;\,\,4 < \left| z \right| < + \infty (7)(z−3)(z−4)(z−1)(z−2)​,3<∣z∣<4;4<∣z∣<+∞

函数 f ( z ) = tan ⁡ ( 1 z ) f\left( z \right) = \tan \left( { {1 \over z}} \right) f(z)=tan(z1​) 能否在 0 < ∣ z ∣ < R   , ( 0 < R < + ∞ ) 0 < \left| z \right| < R\,,\left( {0 < R < + \infty } \right) 0<∣z∣<R,(0<R<+∞) 内展开成洛朗级数？为什么？

  如果 k k k 满足 k 2 < 1 k^2 < 1 k2<1 的实数，证明 ∑ n = 0 ∞ k n sin ⁡ [ ( n + 1 ) θ ] = sin ⁡ θ 1 − 2 k cos ⁡ θ + k 2 \sum\limits_{n = 0}^\infty {k^n \sin \left[ {\left( {n + 1} \right)\theta } \right]} = { {\sin \theta } \over {1 - 2k\cos \theta + k^2 }} n=0∑∞​knsin[(n+1)θ]=1−2kcosθ+k2sinθ​ ∑ n = 0 ∞ k n cos ⁡ [ ( n + 1 ) θ ] = cos ⁡ θ − k 1 − 2 k cos ⁡ θ + k 2 \sum\limits_{n = 0}^\infty {k^n \cos \left[ {\left( {n + 1} \right)\theta } \right]} = { {\cos \theta - k} \over {1 - 2k\cos \theta + k^2 }} n=0∑∞​kncos[(n+1)θ]=1−2kcosθ+k2cosθ−k​

如果 C C C 为正向圆周， ∣ z ∣ = 3 \left| z \right| = 3 ∣z∣=3 。求积分 ∮ C f ( z ) d z \oint_C {f\left( z \right)dz} ∮C​f(z)dz 的值。设 f ( z ) f\left( z \right) f(z) 分别为 ( 1 )      1 z ( z + 2 ) ;                        ( 2 )     z + 2 ( z + 1 ) z ; \left( 1 \right)\,\,\,\,{1 \over {z\left( {z + 2} \right)}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\,\,\,{ {z + 2} \over {\left( {z + 1} \right)z}}; (1)z(z+2)1​;(2)(z+1)zz+2​; ( 3 )      1 z ( z + 1 ) 2 ;                      ( 4 )      z ( z + 1 ) ( z + 2 ) ; \left( 3 \right)\,\,\,\,{1 \over {z\left( {z + 1} \right)^2 }};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\,\,\,\,{z \over {\left( {z + 1} \right)\left( {z + 2} \right)}}; (3)z(z+1)21​;(4)(z+1)(z+2)z​;

● 相关图表链接:

• 图1.1 洛朗级数展开
• 图2.1 洛朗级数收敛域