# How do you solve x^5+1=0?

Mar 10, 2017

Solutions are $\cos \left(\frac{\pi}{5}\right) + i \sin \left(\frac{\pi}{5}\right)$
$- \cos \left(\frac{2 \pi}{5}\right) + i \sin \left(\frac{2 \pi}{5}\right)$, $- 1$
$- \cos \left(\frac{2 \pi}{5}\right) - i \sin \left(\frac{2 \pi}{5}\right)$ and $\cos \left(\frac{\pi}{5}\right) - i \sin \left(\frac{\pi}{5}\right)$

#### Explanation:

As ${x}^{5} + 1 = 0$, we have ${x}^{5} = - 1$ and $x = \sqrt[5]{- 1} = {\left(- 1\right)}^{\frac{1}{5}}$

Hence solution of ${x}^{5} + 1 = 0$ means to find fifth roots of $- 1$.

Note that as $- 1 = \cos \pi + i \sin \pi$, and we can also write

$- 1 = \cos \left(2 n \pi + \pi\right) + i \sin \left(2 n \pi + \pi\right)$

and using De Moivre's Theorem

${\left(- 1\right)}^{\frac{1}{5}} = \cos \left(\frac{2 n \pi + \pi}{5}\right) + i \sin \left(\frac{2 n \pi + \pi}{5}\right)$

and five roots, which are solutions of ${x}^{5} + 1 = 0$ can be obtained by putting $n = 0 , 1 , 2 , 3$ and $4$ (after $4$ roots will start repeating) and these are

$\cos \left(\frac{\pi}{5}\right) + i \sin \left(\frac{\pi}{5}\right)$
$\cos \left(\frac{3 \pi}{5}\right) + i \sin \left(\frac{3 \pi}{5}\right) = - \cos \left(\frac{2 \pi}{5}\right) + i \sin \left(\frac{2 \pi}{5}\right)$
$\cos \left(\frac{5 \pi}{5}\right) + i \sin \left(\frac{5 \pi}{5}\right) = \cos \pi + i \sin \pi = - 1$
$\cos \left(\frac{7 \pi}{5}\right) + i \sin \left(\frac{7 \pi}{5}\right) = - \cos \left(\frac{2 \pi}{5}\right) - i \sin \left(\frac{2 \pi}{5}\right)$
$\cos \left(\frac{9 \pi}{5}\right) + i \sin \left(\frac{9 \pi}{5}\right) = \cos \left(\frac{\pi}{5}\right) - i \sin \left(\frac{\pi}{5}\right)$