Question

计算积分∮c 1/[(z-i）（z+2)]dz ，c为正向圆周｜z｜=3

Answer 1

问：计算积分∮c 1/[(z-i）（z+2)]dz ，c为正向圆周｜z｜=3

答：先分解z+1 A B C----------------- = ------ + ---------- + ---------(z-1)^2*(z+2) z-1 (z-1)^2 z+2z+1=A(z-1)(z+2)+B(z+2)+C(z-1)^2代入z=12=3B, B=2/3代入z=-2-1=9C, C=-1/9代入z=01=-2A+2B+CA=(2B+C-1)/2=1/9原积分=(1/9)∮c[1/(z-1)]dz+(2/3)∮c[1/(z-1)^2]dz-(1/9)∮c[1/(z+2)]dz=(1/9)*2pi*i*1+(2/3)*2pi*i/1! *1-(1/9)*2pi*i*1=(4pi/3)i