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

Oct 23, 2017

$i$, $\text{ "-sqrt(3)/2i-1/2i" }$ and $\text{ } \frac{\sqrt{3}}{2} i - \frac{1}{2} i$

#### Explanation:

The primitive complex cube root of $1$ is:

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

Note that:

${i}^{3} = {i}^{2} \cdot i = - i$

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

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

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

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]}