求f(z)=1/z(z-2)²在0

求f(z)=1/z(z-2)²在0
求f(z)=1/z(z-2)²在0

求f(z)=1/z(z-2)²在0
因为
1/(z-2)^2
= -(1/(z-2))'
= (1/2)(1/(1-(z/2)))'
= (1/2)(求和{n=0,无穷大}(z/2)^n)'
= (1/2)(求和{n=1,无穷大}nz^n-1/2^n)
所以
1/z(z-2)^2 = = 求和{n=1,无穷大}nz^n-2/2^(n+1)