# How do you find the two square roots of -2+2sqrt3i?

Feb 16, 2018

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

#### Explanation:

Note that:

$- 2 + 2 \sqrt{3} i = 4 \omega$

where:

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

is the primitive complex cube root of $1$

So ${\omega}^{3} = 1$ and ${\left({\omega}^{2}\right)}^{2} = {\omega}^{3} \omega = \omega$

That is:

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

$\pm \sqrt{- 2 + 2 \sqrt{3} i} = \pm \sqrt{4} \overline{\omega} = \pm \left(1 + \sqrt{3} i\right)$