# How do you find the 6th root of sqrt3-i ?

Mar 12, 2017

Let $\omega = \sqrt{3} - i$, and we we define $z$ such that ${z}^{6} = \omega$

First let us plot the point $\omega$ on the Argand diagram: And we will put the complex number into polar form:

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

So then in polar form we have:

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

We now want to solve the equation for $z$ (to gain $6$ solutions):

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

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of $2 \pi$, so we can equivalently write (incorporating the periodicity):

${z}^{6} = 2 \left(\cos \left(2 n \pi - \frac{\pi}{6}\right) + i \sin \left(2 n \pi - \frac{\pi}{6}\right)\right) \setminus \setminus \setminus n \in \mathbb{N}$

By De Moivre's Theorem we can write this as:

$z = {\left\{2 \left(\cos \left(2 n \pi - \frac{\pi}{6}\right) + i \sin \left(2 n \pi - \frac{\pi}{6}\right)\right)\right\}}^{\frac{1}{6}}$
$\setminus \setminus = {2}^{\frac{1}{6}} \left(\cos \left(\frac{2 n \pi - \frac{\pi}{6}}{6}\right) + i \sin \left(\frac{2 n \pi - \frac{\pi}{6}}{6}\right)\right)$
$\setminus \setminus = {2}^{\frac{1}{6}} \left(\cos \theta + i \sin \theta\right) \setminus \setminus \setminus \setminus$ where $\theta = \frac{\left(12 n - 1\right) \pi}{36}$

Put:

$n = 1 \implies \theta = \frac{11 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{11 \pi}{36}\right) + i \sin \left(\frac{11 \pi}{36}\right)\right)$

$n = 2 \implies \theta = \frac{23 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{23 \pi}{36}\right) + i \sin \left(\frac{23 \pi}{36}\right)\right)$

$n = 3 \implies \theta = \frac{35 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{35 \pi}{36}\right) + i \sin \left(\frac{35 \pi}{36}\right)\right)$

$n = 4 \implies \theta = \frac{47 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{47 \pi}{36}\right) + i \sin \left(\frac{47 \pi}{36}\right)\right)$

$n = 5 \implies \theta = \frac{59 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{59 \pi}{36}\right) + i \sin \left(\frac{59 \pi}{36}\right)\right)$

$n = 6 \implies \theta = \frac{71 \pi}{36} \implies z = {2}^{\frac{1}{6}} \left(\cos \left(\frac{71 \pi}{36}\right) + i \sin \left(\frac{71 \pi}{36}\right)\right)$

After which the pattern continues.

We can plot these solutions on the Argand Diagram 