# How do you solve 2x^4+2+2sqrt3i=0?

Feb 4, 2017

$x = {2}^{\frac{1}{4}} \left(\cos \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{5 \pi}{12}\right)\right)$ or ${2}^{\frac{1}{4}} \left(- \cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$ or ${2}^{\frac{1}{4}} \left(- \cos \left(\frac{5 \pi}{12}\right) - i \sin \left(\frac{5 \pi}{12}\right)\right)$ or ${2}^{\frac{1}{4}} \left(\cos \left(\frac{\pi}{12}\right) - i \sin \left(\frac{\pi}{12}\right)\right)$

#### Explanation:

As $2 {x}^{4} + 2 + 2 \sqrt{3} i = 0$

${x}^{4} = - 1 - \sqrt{3} i$

Hence $x = {\left(- 1 - \sqrt{3} i\right)}^{\frac{1}{4}}$

To find this we intend to use DeMoivres Theorem. So let us convert $- 1 - \sqrt{3} i$ in polar form.

As $| - 1 - \sqrt{3} i | = 2$, $- 1 - \sqrt{3} i = 2 \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$

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

and $x = {\left(- 1 - \sqrt{3} i\right)}^{\frac{1}{4}} = {2}^{\frac{1}{4}} \left(\cos \left(\frac{2 n \pi}{4} + \frac{5 \pi}{12}\right) + i \sin \left(\frac{2 n \pi}{4} + \frac{5 \pi}{12}\right)\right)$

= ${2}^{\frac{1}{4}} \left(\cos \left(\frac{n \pi}{2} + \frac{5 \pi}{12}\right) + i \sin \left(\frac{n \pi}{2} + \frac{5 \pi}{12}\right)\right)$

Now choosing four values of $n$ i.e. $0 , 1 , 2$ and $3$, we can get four roots of the equation $2 {x}^{4} + 2 + 2 \sqrt{3} i = 0$, which are

${2}^{\frac{1}{4}} \left(\cos \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{5 \pi}{12}\right)\right)$,

${2}^{\frac{1}{4}} \left(\cos \left(\frac{11 \pi}{12}\right) + i \sin \left(\frac{11 \pi}{12}\right)\right)$ i.e. ${2}^{\frac{1}{4}} \left(- \cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$

${2}^{\frac{1}{4}} \left(\cos \left(\frac{17 \pi}{12}\right) + i \sin \left(\frac{17 \pi}{12}\right)\right)$ i.e. ${2}^{\frac{1}{4}} \left(- \cos \left(\frac{5 \pi}{12}\right) - i \sin \left(\frac{5 \pi}{12}\right)\right)$

and ${2}^{\frac{1}{4}} \left(\cos \left(\frac{23 \pi}{12}\right) + i \sin \left(\frac{23 \pi}{12}\right)\right)$ i.e. ${2}^{\frac{1}{4}} \left(\cos \left(\frac{\pi}{12}\right) - i \sin \left(\frac{\pi}{12}\right)\right)$