4数分女.c为1u吗2 上海充通大 解：z=0为被积函数的一阶极点，z=1为二阶极点，而 e Res[f(z),0]=limz·- e G Dyi (-1y =1 z→0 Res[f(),1]= 1 e -lim e(3-0=0. 1 2 所fE二d:=2 rifReq/)0+-Rol/-). =2πi(1+0)=2πi.例： 计算积分 I= 2 ( 1) z C e dz z z − Ñ∫ , C 为正向圆周|z|=2. 1. ( 1) lim ( 1) Res[ ( ),0] lim 2 0 2 0 = − = − = ⋅ → → z e z z e f z z z z z z 2 2 1 1d e Res[ ( ),1] lim ( 1) (2 1)! d ( 1) z z f z z → z zz   = −   − −   2 1 1 d e ( 1) lim lim 0. d z z z z e z → → zz z   − = = =     3 e d 2π {Res[ ( ),0] Res[ ( ),1]} ( 1) 2π (1 0) 2 π . z C z i fz fz z z i i = + − = += 所以∫ 解： z=0为被积函数的一阶极点, z=1为二阶极点, 而