# Find the roots of z^5+1=0?

Apr 12, 2017

Working to 3dp there are 5 solutions:

 z = -1; -0.309+-0.951; z = 0.809+-0.588i

#### Explanation:

Let $\omega = - 1$, and then ${z}^{5} = \omega$

And we will put the complex number into polar form (visually):

$| \omega | = 1$
$a r g \left(\omega\right) = \pi$

So then in polar form we have:

$\omega = \cos \left(\pi\right) + i \sin \left(\pi\right)$

We now want to solve the equation ${z}^{5} = \omega$ for $z$ (to gain $5$ solutions):

${z}^{5} = \cos \left(\pi\right) + i \sin \left(\pi\right)$

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

${z}^{5} = \cos \left(\pi + 2 n \pi\right) + i \sin \left(\pi + 2 n \pi\right) \setminus \setminus \setminus n \in \mathbb{Z}$

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

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

Working to 3dp we get:

Put:

$n = - 2 \implies \theta = - \frac{3 \pi}{5}$
$\text{ } \therefore z = \cos \left(- \frac{3 \pi}{5}\right) + i \sin \left(- \frac{3 \pi}{5}\right)$
$\text{ } \therefore z = - 0.309 - 0.951 i$

$n = - 1 \implies \theta = - \frac{\pi}{5}$
$\text{ } \therefore z = \cos \left(- \frac{\pi}{5}\right) + i \sin \left(- \frac{\pi}{5}\right)$
$\text{ } \therefore z = 0.809 - 0.588 i$

$n = 0 \implies \theta = \frac{\pi}{5}$
$\text{ } \therefore z = \cos \left(\frac{\pi}{5}\right) + i \sin \left(\frac{\pi}{5}\right)$
$\text{ } \therefore z = 0.809 + 0.588 i$

$n = 1 \implies \theta = \frac{3 \pi}{5}$
$\text{ } \therefore z = \cos \left(\frac{3 \pi}{5}\right) + i \sin \left(\frac{3 \pi}{5}\right)$
$\text{ } \therefore z = - 0.309 + 0.951$

$n = 2 \implies \theta = \pi$
$\text{ } \therefore z = \cos \pi + i \sin \pi$
$\text{ } \therefore z = - 1$

After which the pattern continues.

We can plot these solutions on the Argand Diagram

Apr 12, 2017

z_k = e^(i(pi/5+(2kpi)/5) for $k = 0 , 1 , . . , 4$

#### Explanation:

If we express $z$ in polar form, $z = \rho {e}^{i \theta}$ we have that:

${z}^{5} = {\rho}^{5} {e}^{i 5 \theta}$

so:

${z}^{5} = - 1 \implies {\rho}^{5} {e}^{i 5 \theta} = {e}^{i \pi} \implies \left\{\begin{matrix}{\rho}^{5} = 1 \\ 5 \theta = \pi + 2 k \pi\end{matrix}\right.$

We have then:

$\left\{\begin{matrix}\rho = 1 \\ \theta = \frac{\pi}{5} + \frac{2 k}{5} \pi\end{matrix}\right.$

for any $k \in \mathbb{Z}$

We have potentially infinite solutions:

${z}_{k} = {e}^{i \left(\frac{\pi}{5} + \frac{2 k}{5} \pi\right)}$

but we can see that if $j \equiv k \mod 5$, then $j - k = 5 n$ with $n \in \mathbb{Z}$ and we have:

$\frac{\pi}{5} + \frac{2 j \pi}{5} = \frac{\pi}{5} + \frac{2 \left(k + 5 n\right) \pi}{5} = \frac{\pi}{5} + \frac{2 k \pi}{5} + 2 n \pi$

so that:

${z}_{j} = {e}^{i \left(\frac{\pi}{5} + \frac{2 j}{5} \pi\right)} = {e}^{i \left(\frac{\pi}{5} + \frac{2 k}{5} \pi + 2 n \pi\right)} = {e}^{i \left(\frac{\pi}{5} + \frac{2 k}{5} \pi\right)} {e}^{i 2 n \pi} = {z}_{k}$

In conclusion we have five different solutions for $k = 0 , 1 , . . , 4$:

z_0 = e^(i(pi)/5) = 1/4((1+sqrt5)+isqrt(10+sqrt5)))

${z}_{1} = {e}^{i \frac{3 \pi}{5}}$

${z}_{2} = {e}^{i \pi} = - 1$

${z}_{3} = {e}^{i \frac{7 \pi}{5}}$

${z}_{4} = {e}^{i \frac{9 \pi}{5}}$

All this points have module equal to $1$, so they lie on the complex plan on the circle with center the origin and radius $\rho = 1$, and they are the vertices of a regular pentagon inscribed in such circle.