Question

What are the roots of `z^3 = -8i` ?

Answer 1

`2i`, `sqrt(3)-i`, `-sqrt(3)-i`

Explanation:

The primitive complex cube root of `1` is:

`omega = -1/2+sqrt(3)/2i`

If `alpha` is one root of `z^3=-8i` then the other two are `omega alpha` and `omega^2 alpha`.

Note that:

`(2i)^3 = 2^3i^3 = -8i`

So `2i` is one root.

So the other roots are:

`omega * 2i = (-1/2+sqrt(3)/2i) 2i = -sqrt(3)-i`

`omega^2 * 2i = (-1/2-sqrt(3)/2i) 2i = sqrt(3)-i`

Here are the three roots in the complex plane, plotted with the circle `abs(z) = 2`:

graph{(x^2+(y-2)^2-0.01)((x-sqrt(3))^2+(y+1)^2-0.01)((x+sqrt(3))^2+(y+1)^2-0.01)(x^2+y^2-4) = 0 [-5, 5, -2.5, 2.5]}