求将函数展开成z的幂级数

Question

如题所述

Answer 1

根据欧拉公式：e^(iθ)=cosθ+i*sinθ
f(z)=z/(z^2+i)
=z/[z^2-(-i)]
=z/[z^2-e^(-iπ/2)]
=z/[z+e^(-iπ/4)][z-e^(-iπ/4)]
=(1/2)*{1/[z+e^(-iπ/4)]+1/[z-e^(-iπ/4)]}
=(1/2)*{1/[e^(-iπ/4)+z]-1/[e^(-iπ/4)-z]}
=[e^(iπ/4)/2]*{1/[1+ze^(iπ/4)]-1/[1-ze^(iπ/4)]}
=[e^(iπ/4)/2]*{∑(n=-∞->∞) [-ze^(iπ/4)]^n-∑(n=-∞->∞) [ze^(iπ/4)]^n}
=[e^(iπ/4)/2]*{∑(n=-∞->∞) (-1)^n*e^(niπ/4)*z^n-∑(n=-∞->∞) e^(niπ/4)*z^n}
=[e^(iπ/4)/2]*∑(n=-∞->∞) [(-1)^n-1]*e^(niπ/4)*z^n
=(1/2)*∑(n=-∞->∞) [(-1)^n-1]*e^[(n+1)iπ/4]*z^n
=∑(n=-∞->∞) -e^(niπ/2)*z^(2n-1)
=∑(n=-∞->∞) e^[(n/2+1)iπ]*z^(2n-1)本回答被提问者和网友采纳