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

Apr 15, 2017

$2 i$, $\sqrt{3} - i$, $- \sqrt{3} - i$

#### Explanation:

The primitive complex cube root of $1$ is:

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

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

Note that:

${\left(2 i\right)}^{3} = {2}^{3} {i}^{3} = - 8 i$

So $2 i$ is one root.

So the other roots are:

$\omega \cdot 2 i = \left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right) 2 i = - \sqrt{3} - i$

${\omega}^{2} \cdot 2 i = \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) 2 i = \sqrt{3} - i$

Here are the three roots in the complex plane, plotted with the circle $\left\mid z \right\mid = 2$:

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]}