Question

在复数范围内-1,1的6个六次方根

求详细过程

Answer 1

设根为z=xe^(iy)
则z^6=-1 和z^6=1
(1) z^6=(x^6)e^(6iy)=-1=e^(iπ+i2nπ)
x^6=1 ---> x=+/-1
6y=(2n+1)π ----> y=(1/6+n/3)π=-1/6π, π/6, π/2, 5π/6, 7π/6, 3π/2
z=xe^(iy)=x(cos(y)+isin(y))=+/-(√3-i)/2, +/-(√3+i)/2, +/-i
(2)z^6=(x^6)e^(6iy)=1=e^(i2nπ)
x^6=1 ---> x=+/-1
6y=2nπ ----> y=nπ/3=-π/3, 0, π/3, 2π/3, π, 4π/3, 5π/3
z=xe^(iy)=x(cos(y)+isin(y))=+/-(1+i√3)/2, +/-(1-√3)/2, +/-1