# A question of Roots of unity

By considering the ninth roots of unity, show that: $$\cos(\frac{2\pi}{9}) + \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$$.

I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $$4$$ roots.

## 3 Answers

Note that$$\begin{multline}\cos\left(\frac{2\pi}9\right)+\cos\left(\frac{4\pi}9\right)+\cos\left(\frac{6\pi}9\right)+\cos\left(\frac{8\pi}9\right)=\\=\frac12\left(e^{2\pi i/9}+e^{-2\pi i/9}+e^{4\pi i/9}+e^{-4\pi i/9}+e^{6\pi i/9}+e^{-6\pi i/9}+e^{8\pi i/9}+e^{-8\pi i/9}\right)\end{multline}$$But this is half the sum of all ninth roots of unity other than $$1$$. So, it's half the sum of the roots of$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x$$and that sum is $$-1$$.

• How do you find the sum of the roots of unity?
– newm
Commented Dec 6, 2020 at 12:38
• Let $\omega$ be a root of unity: i.e $\omega^9 = 1$. Then $1 + \omega + \omega^2 + \cdots + \omega^8 = \frac{1 - \omega^9}{1 - \omega} = 0$ using the sum for a finite geometric series. Subtracting $1$ gives $\omega + \omega^2 + \cdots + \omega^8 = -1$. Commented Dec 27, 2020 at 8:13
• @JoséCarlosSantos You have a typo: there should not be a $1$ in $x^8 + x^7 + \cdots + 1$. Commented Dec 27, 2020 at 8:14

The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $$e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$$

Dividing both sides $$z^4,$$

$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$

$$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$$(say)

Use $$\left(z+\dfrac1z\right)^2=z^2+\dfrac1{z^2}+2$$

$$\left(z+\dfrac1z\right)^3=z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$$

$$z^4+\dfrac1{z^4}=\left(z^2+\dfrac1{z^2}\right)^2-2=\left(\left(z+\dfrac1z\right)^2-2\right)^2-2$$

to form a bi-quadratic equation in $$y$$

Now use Vieta's formula

The roots of unity for $$z^9=1$$ are $$e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$$ and their sum is equal to zero, i.e. $$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9} + e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9} + e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9} + e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9}$$ which leads to $$\cos\frac{2\pi}{9}+ \cos\frac{4\pi}{9}+ \cos\frac{6\pi}{9} + \cos\frac{8\pi}{9} =- \frac{1}{2}$$