Question

解关于复数z方程z^3+8i=0

Answer 1

z^3+8i=0

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

扩展资料

一个数有多少个方根，这个问题既与数的所在范围有关，也与方根的次数有关。在实数范围内，任一实数的奇数次方根有且仅有一个，例如8的3次方根为2，-8的 3次方根为-2。

正实数的偶数次方根是两个互为相反数的数，例如16的4次方根为2和-2；负实数不存在偶数次方根；零的任何次方根都是零。在复数范围内，无论n是奇数或偶数，任一个非零的复数的n次方根都有n个。如果复数

r=z，那么它的n个n次方根是，k=0，1，2…，n-1。

Answer 2

z^3+8i=0
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