# How do you use DeMoivre's theorem to simplify (-sqrt3-i)^4?

Sep 10, 2016

$- 8 + 8 \sqrt{3} i$

#### Explanation:

We have: ${\left(- \sqrt{3} - i\right)}^{4}$

First, let's consider the complex number $z = - \sqrt{3} - i$.

In order to apply De Moivre's theorem, we need to evaluate the modulus and argument of this $z$:

=> |z| = sqrt((- sqrt(3)^(2) + (- 1)^(2))

$\implies | z | = \sqrt{3 + 1}$

$\implies | z | = \sqrt{4}$

$\implies | z | = 2$

$\implies \theta = \arctan \left(\frac{- 1}{- \sqrt{3}}\right)$

$\implies \theta = \arctan \left(\frac{\sqrt{3}}{3}\right)$

$\implies \theta = \frac{\pi}{6}$

Then, $z$ is located in the third quadrant:

$R i g h t a r r o w a r g \left(z\right) = \frac{\pi}{6} - \pi = - \frac{5 \pi}{6}$

So, $z = 2 \left(\cos \left(- \frac{5 \pi}{6}\right) + i \sin \left(- \frac{5 \pi}{6}\right)\right)$

Now, using De Moivre's theorem:

$\implies {z}^{4} = {2}^{4} \left(\cos \left(4 \cdot - \frac{5 \pi}{6}\right) + i \sin \left(4 \cdot - \frac{5 \pi}{6}\right)\right)$

$\implies {z}^{4} = 16 \left(\cos \left(- \frac{10 \pi}{3}\right) + i \sin \left(- \frac{10 \pi}{3}\right)\right)$

$\implies {z}^{4} = 16 \left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

$\implies {z}^{4} = 16 \left(- \frac{1}{2} \left(1 - \sqrt{3} i\right)\right)$

$\implies {z}^{4} = - 8 \left(1 - \sqrt{3} i\right)$

$\therefore {z}^{4} = - 8 + 8 \sqrt{3} i$

Therefore, ${\left(- \sqrt{3} - i\right)}^{4} = - 8 + 8 \sqrt{3} i$.