# How do you find all the real and complex roots of z^5 + 1 = 0?

Mar 16, 2018

$z = - 1$

$z = \frac{1}{4} \left(\sqrt{5} - 1\right) \pm \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} i$

$z = \frac{1}{4} \left(- \sqrt{5} - 1\right) \pm \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} i$

#### Explanation:

$0 = {z}^{5} + 1$

$\textcolor{w h i t e}{0} = \left(z + 1\right) \left({z}^{4} - {z}^{3} + {z}^{2} - z + 1\right)$

$\textcolor{w h i t e}{0} = \left(z + 1\right) {z}^{2} \left({z}^{2} - z + 1 - \frac{1}{z} + \frac{1}{z} ^ 2\right)$

$\textcolor{w h i t e}{0} = \left(z + 1\right) {z}^{2} \left({\left(z + \frac{1}{z}\right)}^{2} + \left(z + \frac{1}{z}\right) - 1\right)$

$\textcolor{w h i t e}{0} = \left(z + 1\right) {z}^{2} \left({\left(\left(z + \frac{1}{z}\right) + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{0} = \left(z + 1\right) {z}^{2} \left(z + \frac{1}{z} + \frac{1}{2} - \frac{\sqrt{5}}{2}\right) \left(z + \frac{1}{z} + \frac{1}{2} + \frac{\sqrt{5}}{2}\right)$

$\textcolor{w h i t e}{0} = \left(z + 1\right) \left({z}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) z + 1\right) \left({z}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) z + 1\right)$

So:

$0 = z + 1 \rightarrow z = - 1$

or:

$0 = {z}^{2} + \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) z + 1$

and hence:

$z = \frac{1}{2} \left(- \frac{1}{2} + \frac{\sqrt{5}}{2} \pm \sqrt{{\left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right)}^{2} - 4}\right)$

$\textcolor{w h i t e}{z} = \frac{1}{4} \left(\sqrt{5} - 1\right) \pm \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} i$

or:

$0 = {z}^{2} + \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) z + 1$

and hence:

$z = \frac{1}{2} \left(- \frac{1}{2} - \frac{\sqrt{5}}{2} \pm \sqrt{{\left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right)}^{2} - 4}\right)$

$\textcolor{w h i t e}{z} = \frac{1}{4} \left(- \sqrt{5} - 1\right) \pm \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} i$