z^3=-8i
因此z=2*(-i 开三次方)
-i 的幅角为3π/2
因此z=2[cos(3π/2+2kπ)/3+isin(3π/2+2kπ)/3]
=2[cos(π/2+2kπ/3)+isin(π/2+2kπ/3)] k=0,1,2
k=0时：z1=2[cos(π/2)+isin(π/2)]=2i
k=1时：z2=2[cos(π/2+2π/3)+isin(π/2+2π/3)]
=2[-sin(2π/3)+icos(2π/3)]
=-√3+i
k=2时：z3=2[cos(π/2+4π/3)+isin(π/2+4π/3)]
=2[-sin(4π/3)+icos(4π/3)]
=2[sin(π/3)-icos(π/3)]
=√3-i请问-i的幅角怎么得来的x轴正向的辐角为0y轴正向的辐角为π/2x轴负向的辐角为πy轴负向的辐角为3π/2-i 在y轴负向上。