# How do you find the power (sqrt3-i)^3 and express the result in rectangular form?

Mar 7, 2017

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

#### Explanation:

Let $z = \sqrt{3} - i$

First let us plot the point $z$ on the Argand diagram:

And we will put the complex number into polar form:

$| z | = \sqrt{3 + 1} = 2$
$a r g \left(z\right) = {\tan}^{-} 1 \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6}$

So then in polar form we have:

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

By De Moivre's Theorem:

${z}^{3} = {\left\{2 \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right)\right\}}^{3}$
$\setminus \setminus \setminus = {2}^{3} \left(\cos \left(- \frac{3 \pi}{6}\right) + i \sin \left(- \frac{3 \pi}{6}\right)\right)$
$\setminus \setminus \setminus = 8 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$
$\setminus \setminus \setminus = 8 \left(0 - i\right)$
$\setminus \setminus \setminus = - 8 i$