# How do you solve 2x^3+4+2i=0?

Dec 23, 2016

The roots are:

${x}_{1} = \sqrt[3]{2 + i}$

${x}_{2} = \omega \sqrt[3]{2 + i}$

${x}_{3} = {\omega}^{2} \sqrt[3]{2 + i}$

#### Explanation:

Given:

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

Divide through by $2$ to get:

${x}^{3} + 2 + i = 0$

Subtract $2 + i$ from both sides to get:

${x}^{3} = 2 + i$

This has roots:

${x}_{1} = \sqrt[3]{2 + i}$

${x}_{2} = \omega \sqrt[3]{2 + i}$

${x}_{3} = {\omega}^{2} \sqrt[3]{2 + i}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

If you prefer these roots in $a + b i$ form, then it can be done with trigonometric functions:

First note that:

$\left\mid 2 + i \right\mid = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$

$A r g \left(2 + i\right) = {\tan}^{- 1} \left(\frac{1}{2}\right)$

Hence, using de Moivre's formula:

${x}_{1} = \sqrt[6]{5} \cos \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right)\right) + i \sqrt[6]{5} \sin \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right)\right)$

${x}_{2} = \sqrt[6]{5} \cos \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right) + \frac{2 \pi}{3}\right) + i \sqrt[6]{5} \sin \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right) + \frac{2 \pi}{3}\right)$

${x}_{3} = \sqrt[6]{5} \cos \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right) + \frac{4 \pi}{3}\right) + i \sqrt[6]{5} \sin \left(\frac{1}{3} {\tan}^{- 1} \left(\frac{1}{2}\right) + \frac{4 \pi}{3}\right)$