# 14.8 The roots of unity

The nth roots of unity are the nth roots of one. How can there be more than one ($1$) nth root of one? Well, some of them are complex, of course!

Note that ^{8}^{8}
8
In this document the natural numbers include zero!

This is because of Euler’s formula.

Therefore, if we take the $n$th roots of both sides, we get that

Which we can use to compute the $n$th roots of unity. Note that by the fundamental theorem of algebra for the $nth$ roots of unity, there are $n$ different values.

Note that we often use the letter $\omega$ (the Greek “omega”) to denote $e^{2\pi/n}$.

###### Theorem 14.8.1

The $n$th roots of unity sum to $\mathrm{0}$.

Proof: I don’t think you need to know this for A Level Mathematics, but an easy-ish proof is to consider

and note that if we multiply through by $\omega_{n}$ we actually get the same value back! That is

$\displaystyle\omega_{n}(1+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-2}+% \omega_{n}^{n-1})$ | $\displaystyle=\omega_{n}+\omega_{n}^{2}+\omega_{n}^{3}+...+\omega_{n}^{n-1}+% \omega_{n}^{n}$ | (14.82) | ||

$\displaystyle=\omega_{n}+\omega_{n}^{2}+\omega_{n}^{3}+...+\omega_{n}^{n-1}+1$ | (14.83) | |||

$\displaystyle=1+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-2}+\omega_{n}^{n-1}$ | (14.84) |

The only complex number for which $ax=x$ is $0$ and therefore the sum of the roots of unity is $0$.

###### Example 14.8.1

By considering the ninth roots of unity, show that
^{9}^{9}
9
I believe this question comes from the textbook
”Further Pure Mathematics”

Solution

Using Euler’s formula
^{10}^{10}
10
$e^{\theta i}=\cos(\theta)+i\sin(\theta)$, see above for more
we can write the sum of
$\cos(\frac{2\pi}{9})+\cos(\frac{4\pi}{9})+\cos(\frac{6\pi}{9})+\cos(\frac{8\pi%
}{9})$ as

This is actually the sum of eight of the nine roots of unity in disguise! Note that if we add $2\pi$ to anything in the form $e^{ai}$, this has no effect (again, Euler’s formula and the fact that $2\pi$ radians is a full rotation). Therefore, the previous expression is the same as

which can be re-ordered as

which are all the ninth roots of unity (except $1$).

Thus we can write that

$\displaystyle\frac{1}{2}\left[\cos\left(\frac{2\pi}{9}\right)+\cos\left(\frac{% 4\pi}{9}\right)+\cos\left(\frac{6\pi}{9}\right)+\cos\left(\frac{8\pi}{9}\right% )\right]$ | $\displaystyle=-1+1+\omega+\omega^{2}+...+\omega^{8}$ | ||

$\displaystyle=-1+\frac{1-\omega^{9}}{\omega-1}$ | |||

$\displaystyle=-1+\frac{1-1}{\omega-1}\text{ ($\omega^{9}=1$ as $\omega$ is a 9% th root of 1)}$ | |||

$\displaystyle=-1$ |

which means that