# 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 88 8 In this document the natural numbers include zero!

$e^{\pm 2n\pi i}=1\text{ where }n\in\mathbb{N}$

This is because of Euler’s formula.

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

$e^{\pm\frac{2\pi}{n}i}=1^{\frac{1}{n}}\text{ where }n\in\mathbb{N}$

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 $0$.

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

$1+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-1}$ (14.81)

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 99 9 I believe this question comes from the textbook ”Further Pure Mathematics”

$\cos\left(\frac{2\pi}{9}\right)+\cos\left(\frac{4\pi}{9}\right)+\cos\left(% \frac{6\pi}{9}\right)+\cos\left(\frac{8\pi}{9}\right)=-\frac{1}{2}$

Solution

Using Euler’s formula 1010 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

$\frac{1}{2}\left[e^{\frac{2\pi}{9}i}+e^{-\frac{2\pi}{9}i}+e^{\frac{4\pi}{9}i}+% e^{-\frac{4\pi}{9}i}+e^{\frac{6\pi}{9}i}+e^{-\frac{6\pi}{9}i}+e^{\frac{8\pi}{9% }i}+e^{-\frac{8\pi}{9}i}\right]$

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

$\frac{1}{2}\left[e^{\frac{2\pi}{9}i}+e^{\frac{16\pi}{9}i}+e^{\frac{4\pi}{9}i}+% e^{\frac{14\pi}{9}i}+e^{\frac{6\pi}{9}i}+e^{\frac{12\pi}{9}i}+e^{\frac{8\pi}{9% }i}+e^{\frac{10\pi}{9}i}\right]$

which can be re-ordered as

$\frac{1}{2}\left[e^{\frac{2\pi}{9}i}+e^{\frac{4\pi}{9}i}+e^{\frac{6\pi}{9}i}+e% ^{\frac{8\pi}{9}i}+e^{\frac{10\pi}{9}i}+e^{\frac{12\pi}{9}i}+e^{\frac{14\pi}{9% }i}+e^{\frac{16\pi}{9}i}\right]$

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

Thus we can write that

which means that

$\cos\left(\frac{2\pi}{9}\right)+\cos\left(\frac{4\pi}{9}\right)+\cos\left(% \frac{6\pi}{9}\right)+\cos\left(\frac{8\pi}{9}\right)=-\frac{1}{2}$