# Question 82567

Mar 24, 2016

$\cos \left(\frac{2 \pi}{9}\right) + i \sin \left(\frac{2 \pi}{9}\right)$, $\cos \left(\frac{8 \pi}{9}\right) + i \sin \left(\frac{8 \pi}{9}\right)$ and

$\cos \left(\frac{14 \pi}{9}\right) + i \sin \left(\frac{14 \pi}{9}\right)$,

#### Explanation:

The first thing to do is to put the number in the form of $\rho {e}^{\theta i}$

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

$\theta = \arctan \left(\frac{\frac{\sqrt{3}}{2}}{- \frac{1}{2}}\right) = \arctan \left(- \sqrt{3}\right) = - \frac{\pi}{3} + k \pi$. Let's choose $\frac{2 \pi}{3}$since we are in the second quadrant. Pay attention that $- \frac{\pi}{3}$ is in the fourth quadrant, and this is wrong.

$1 {e}^{\frac{2 \pi i}{3}}$

Now the roots are:

$\sqrt[3]{1} {e}^{\frac{\left(2 k \pi + \frac{2 \pi}{3}\right) i}{3}} , k \in \mathbb{Z}$

=e^((((6kpi+2pi)i)/9), k in ZZ

so you can choose k= 0, 1, 2 and obtain:

e^((2pii)/9, e^((8kpii)/9 and e^((14kpii)/9#

or $\cos \left(\frac{2 \pi}{9}\right) + i \sin \left(\frac{2 \pi}{9}\right)$, $\cos \left(\frac{8 \pi}{9}\right) + i \sin \left(\frac{8 \pi}{9}\right)$ and

$\cos \left(\frac{14 \pi}{9}\right) + i \sin \left(\frac{14 \pi}{9}\right)$.

For me this is a dead end, because I cannot compute trigonometric functions of multiples of $\frac{\pi}{9}$. We must rely on a calculator:

$0.7660 + 0.6428 i$
$- 0.9397 + 0.3420 i$
$0.1736 - 0.9848 i$