复变函数题目,有会的同学吗?不是考试。麻烦帮我做一下,我要解题过程,麻烦帮帮忙,谢谢大家了!

2025-04-14 03:24:31
推荐回答(1个)
回答1:

(1)
f(z)=1/(z-1)·1/(z-2)
=1/(z-1)·1/(z-1-1)
=-1/(z-1)·1/[1-(z-1)]
=-1/(z-1)·[1+(z-1)+(z-1)²+(z-1)³+……]
=-1/(z-1)-1-(z-1)-(z-1)²-……

(2)
f(z)=1/(z-2)·1/(z-1)
=1/(z-2)·1/(z-2+1)
=1/(z-2)²·1/[1+1/(z-2)]
=1/(z-2)²·[1-1/(z-2)+1/(z-2)²-1/(z-2)³+……]
=1/(z-2)²-1/(z-2)³+1/(z-2)^4-1/(z-2)^5+……