z^3=-8i
因此z=2*(-i 開三次方)
-i 的幅角為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時(shí)：z1=2[cos(π/2)+isin(π/2)]=2i
k=1時(shí)：z2=2[cos(π/2+2π/3)+isin(π/2+2π/3)]
=2[-sin(2π/3)+icos(2π/3)]
=-√3+i
k=2時(shí)：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請(qǐng)問-i的幅角怎么得來(lái)的x軸正向的輻角為0y軸正向的輻角為π/2x軸負(fù)向的輻角為πy軸負(fù)向的輻角為3π/2-i 在y軸負(fù)向上。