Question

将函数fz=1/z(1-z)^2在圆环0<‖z-1‖<1,及1<‖z-1‖<正无穷内展开洛朗级数

Answer 1

f(z)=-1/z*1/(z-1)²
其中1/(z-1)²已经是z=1这一点的展开了，所以只要考虑把-1/z在z=1这一点展开成洛朗级数。
当0<|z-1|<1时，-1/z=-1/(1+(z-1))=自己写
当1<|z-1|<+∞时，有0<1/(z-1)<1，-1/z=-1/(z-1)*1/(1+(1/(z-1))=自己写

Answer 2

设
f(z)=A/z+B/(1-z)+C/(1-z)²
=[A(1-z)²+Bz（1-z）+Cz]/z(1-z)²
=[A+(-2A+B+C)z+(A-B)z²]/z(1-z)²
A=1
-2A+B+C=0
A-B=0
B=A=1
C=2A-B=A=1
f(z)=1/z+1/(1-z)+1/(1-z)²
对于0<||Z-1||＜1
f(z)=1/z+1/(1-z)+1/(1-z)²
=1/(z-1+1)+1/(1-z)+1/(1-z)²
=1/[1-(1-z)]+1/(1-z)+1/(1-z)²
=1/(1-z)²+1/(1-z)+1+(1-z)+(1-z)²+...+(1-z)^k+.....
对于1<||z-1||<+∞
f(z)=1/z+1/(1-z)+1/(1-z)²
=f(z)=1/（z-1+1）-1/(z-1)+1/(z-1)²
=[1/(z-1)]/[1+1/(z-1)]-1/(z-1)+1/(z-1)²
=....（-1）^(k+1)/(z-1)^k+......+1/(z-1)³-1/（z-1）²+1/(z-1)-1/(z-1)+1/(z-1)²
=....（-1）^(k+1)/(z-1)^k+......-1/(z-1)^4+1/(z-1)³