How do you find the square root of 16 ( cos ((2pi)/3) + i sin ((2pi)/3))?

Jul 5, 2016

$\sqrt{16 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)} = 4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) = 2 + 2 \sqrt{3} i$

Explanation:

If $r \ge 0$ and $- \pi < \theta \le \pi$ then:

$\sqrt{r \left(\cos \theta + i \sin \theta\right)} = \sqrt{r} \left(\cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right)\right)$

So in our example:

$\sqrt{16 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)}$

$= 4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$

$= 4 \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

$= 2 + 2 \sqrt{3} i$

Note that this is the principal square root.

The other square root is:

$- 4 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) = - 2 - 2 \sqrt{3} i$