Question

How do you find three cube roots of `-i`?

Answer 1

`i`, `" "-sqrt(3)/2i-1/2i" "` and `" "sqrt(3)/2i-1/2i`

Explanation:

The primitive complex cube root of `1` is:

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

Note that:

`i^3 = i^2 * i = -i`

So one of the cube roots is `i`. The other two cube roots can be found by multiplying by powers of `omega`.

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

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

Here are the three cube roots of `-i` plotted in the complex plane, together with the unit circle on which they lie...

graph{(x^2+(y-1)^2-0.002)((x-sqrt(3)/2)^2+(y+1/2)^2-0.002)((x+sqrt(3)/2)^2+(y+1/2)^2-0.002)(x^2+y^2-1) = 0 [-2.5, 2.5, -1.25, 1.25]}