4.1.留数定理 1/38 3.推论二 设P(z),Q(z)在点z0解析,且 P(z0)≠0,Q(z0)=0,Q(x0)≠0(即一阶零点),如果z0是函数 f(z)=P(z)/Q(z)的一阶极点,则 Resf(zo)= lim(z-z0) P(x)1P(z0) (4.1-13) 2(z) o 由推论1可以立即得证 (z-z0)P(z) P(z)+(z-zo)P(z) P( Im E im x→0 2() z→20 e(z) Q(zo 413例题 S例1求f(x)=1/(x-1在z0=1的留数 ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §4.1. 3ê½n 11/38 3. íØ . P(z), Q(z) 3. :. z0 ). Û. § . P(z0) , 0§Q(z0) = 0§Q 0 (z0) , 0£=. . . ". :. ¤§X. J. z0 ´. ¼. ê. f(z) = P(z)/Q(z) . . . 4. :. §K. ~ Resf(z0) = lim z→z0 (z − z0) P(z) Q(z) = P(z0) Q0 (z0) . (4.1-13) díج±á=y© lim z→z0 (z − z0)P(z) Q(z) = lim z→z0 P(z) + (z − z0)P 0 (z) Q0 (z) = P(z0) Q0 (z0) 4.1.3 ~K ✿ ~¬¦ f(z) = 1/(z n − 1) 3 z0 = 1 3ê©
4.1.留数定理 12/38 解:由于lim2→1f(z)=∞,因此z0=1是函数的极点.又因为 (利用(41-1) lim (z-1) lim 21|zn-1+z-1+…+z+1 所以z0=1是单极点.由单极点留数计算式(41-12)得 Resf(1)= lim (z-1) 1z1z-1+z"-1+…+z+1 另外,最简单的办法是使用(41-13 Res(1)= n2n-1|x=20 S例2确定函数f(x)=1/sinz的极点,求出函数在这些极点的留 数 解:由于z→m(n为整数,包括零),有sinz→0,f(z)→∞, 因此,z0=n是极点 Z- nt lm[x-丌)j(z z→n z→n SIn z ● First●Prev●Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §4.1. 3ê½n 12/38 )µdu limz→1 f(z) = ∞§Ïd z0 = 1 ´¼ê4:©qÏ £|^(4.1-1)¤ lim z→1 (z − 1) 1 z n − 1 = lim z→1 1 z n−1 + z n−1 + · · · + z + 1 = 1 n ¤± z0 = 1 ´ü4:©dü4:3êOª(4.1-12) Resf(1) = lim z→1 (z − 1) 1 z n − 1 = lim z→1 1 z n−1 + z n−1 + · · · + z + 1 = 1 n , §{ü{´¦^(4.1-13)§ Resf(1) = 1 (z n − 1)0 z=z0 = 1 nzn−1 z=z0 = 1 n . ✿ ~(½¼ê f(z) = 1/ sin z 4:§¦Ñ¼ê3ù 4:3 ê© )µdu z → nπ (n ê§)")§k sin z → 0, f(z) → ∞§ Ïd§z0 = nπ ´4:© lim z→nπ [(z − nπ)f(z)] = lim z→nπ z − nπ sin z
4.1.留数定理 13/38 利用罗毕达法则,有 lim[(z-nx)f()=mim、 (-1) z→n → n COS Z 所以z0=mx是单极点函数f(xz)在单极点的留数等于 Resf(n)= lim[(z-n)f(]=(1) z→n S例3确定函数f(x)=(x+2i)/(z5+4x3)的极点,并求出函数在这些 极点的留数 解:先化简函数 2i z+ 21 f(z) (z2+4)z3(z+2i)(z-2i)x3(z-2i) 显然,z0=2i和z0=0是函数f(x)的极点因为z0=2和z0=0分 别是分母z(n-2i)一阶和三零点,所以z0=2i和z0=0分别是函数 的单极点和三阶极点.由留数计算式(?),有 Re(2)=妈m lim x3(z-2i)z-2z3-8i8 ●Fist●Prev·Next●Last● Go Back● Full Screen●cose●Quit
• First • Prev • Next • Last • Go Back • Full Screen • Close • Quit §4.1. 3ê½n 13/38 |^Û.{K§k lim z→nπ [(z − nπ)f(z)] = lim z→nπ 1 cos z = (−1)n . ¤±z0 = nπ ´ü4:©¼ê f(z) 3ü4:3ê u Resf(nπ) = lim z→nπ [(z − nπ)f(z)] = (−1)n . ✿ ~®(½¼ê f(z) = (z + 2i)/(z 5 + 4z 3 ) 4:§¿¦Ñ¼ê3ù 4:3ê© )µkz{¼ê f(z) = z + 2i z 3 (z 2 + 4) = z + 2i z 3 (z + 2i)(z − 2i) = 1 z 3 (z − 2i). w,§z0 = 2i Ú z0 = 0 ´¼ê f(z) 4:©Ïz0 = 2i Ú z0 = 0 © O´©1 z 3 (n − 2i) Ún":§¤±z0 = 2i Ú z0 = 0 ©O´¼ê ü4:Ún 4:©d3êOª(??) §k Resf(2i) = lim z→2i (z − 2i) 1 z 3 (z − 2i) = lim z→2i 1 z 3 = 1 −8i = i 8 ,