# How do you find the fourth root of 16(cos((4pi)/3)+isin((4pi)/3))?

Nov 20, 2016

$\sqrt[4]{16 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)} = 2 \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right) = \sqrt{3} - i$

#### Explanation:

De Moivre's formula tells us that:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

So we find:

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

$\textcolor{w h i t e}{{\left(2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)\right)}^{4}} = 16 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)$

So $2 \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) = 1 + \sqrt{3} i$ is a fourth root of $16 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)$

Is it the fourth root?

No.

Note that $\frac{4 \pi}{3}$ is in Q3. So if we are using the normal range for $A r g \left(z\right) \in \left(- \pi , \pi\right]$ then the standard form of $\frac{4 \pi}{3}$ is $\frac{- 2 \pi}{3}$ and the principal fourth root is:

$2 \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right) = \sqrt{3} - i$

which is in Q4.

Here are the four fourth roots of $16 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)$ plotted in the Complex plane...

graph{((x-1)^2+(y-sqrt(3))^2-0.008)((x-sqrt(3))^2+(y+1)^2-0.008)((x+1)^2+(y+sqrt(3))^2-0.008)((x+sqrt(3))^2+(y-1)^2-0.008) = 0 [-5, 5, -2.5, 2.5]}