复变函数题目，有会的同学吗
职业培训
培训职业
2024-12-15
(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)²1/[1+1/(z-2)]=1/(z-2)²[1-1/(z-2)+1/(z-2)²-1/(z-2)
(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+……
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