# Find the argument of (-sqrt3 +i)^8?

Apr 2, 2017

Argument is $\frac{2 \pi}{3}$

#### Explanation:

${\left(- \sqrt{3} + i\right)}^{8}$

= ${\left[2 \left(- \frac{\sqrt{3}}{2} + i \times \frac{1}{2}\right)\right]}^{8}$

= ${\left[2 \left(- \frac{\sqrt{3}}{2} + i \times \frac{1}{2}\right)\right]}^{8}$

= ${\left[2 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)\right]}^{8}$

Now according to DeMoivre's Theorem

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

Hence ${\left[2 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)\right]}^{8}$

= $\left[{2}^{8} \left(\cos \left(\frac{5 \pi}{6} \times 8\right) + i \sin \left(\frac{5 \pi}{6} \times 8\right)\right)\right]$

= $256 \left(\cos \left(\frac{20 \pi}{3}\right) + i \sin \left(\frac{20 \pi}{3}\right)\right)$

= $256 \left(\cos \left(6 \pi + \frac{2 \pi}{3}\right) + i \sin \left(6 \pi + \frac{2 \pi}{3}\right)\right)$

= $256 \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$

Hence while modulus of ${\left(- \sqrt{3} + i\right)}^{8}$ is $256$, argument is $\frac{2 \pi}{3}$