14.8 The roots of unity

The nth roots of unity are the nth roots of one. How can there be more than one (11) 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±2⁢n⁢π⁢i=1⁢ where ⁢n∈ℕe^{\pm 2n\pi i}=1\text{ where }n\in\mathbb{N}

This is because of Euler’s formula.

Therefore, if we take the nnth roots of both sides, we get that

e±2⁢πn⁢i=11n⁢ where ⁢n∈ℕe^{\pm\frac{2\pi}{n}i}=1^{\frac{1}{n}}\text{ where }n\in\mathbb{N}

Which we can use to compute the nnth roots of unity. Note that by the fundamental theorem of algebra for the n⁢t⁢hnth roots of unity, there are nn different values.

Note that we often use the letter ω\omega (the Greek “omega”) to denote e2⁢π/ne^{2\pi/n}.

Theorem 14.8.1

The nnth 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+ωn+ωn2+…+ωnn−11+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-1} (14.81)

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

ωn⁢(1+ωn+ωn2+…+ωnn−2+ωnn−1)\displaystyle\omega_{n}(1+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-2}+% \omega_{n}^{n-1}) =ωn+ωn2+ωn3+…+ωnn−1+ωnn\displaystyle=\omega_{n}+\omega_{n}^{2}+\omega_{n}^{3}+...+\omega_{n}^{n-1}+% \omega_{n}^{n} (14.82)
=ωn+ωn2+ωn3+…+ωnn−1+1\displaystyle=\omega_{n}+\omega_{n}^{2}+\omega_{n}^{3}+...+\omega_{n}^{n-1}+1 (14.83)
=1+ωn+ωn2+…+ωnn−2+ωnn−1\displaystyle=1+\omega_{n}+\omega_{n}^{2}+...+\omega_{n}^{n-2}+\omega_{n}^{n-1} (14.84)

The only complex number for which a⁢x=xax=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⁡(2⁢π9)+cos⁡(4⁢π9)+cos⁡(6⁢π9)+cos⁡(8⁢π9)=−12\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 eθ⁢i=cos⁡(θ)+i⁢sin⁡(θ)e^{\theta i}=\cos(\theta)+i\sin(\theta), see above for more we can write the sum of cos⁡(2⁢π9)+cos⁡(4⁢π9)+cos⁡(6⁢π9)+cos⁡(8⁢π9)\cos(\frac{2\pi}{9})+\cos(\frac{4\pi}{9})+\cos(\frac{6\pi}{9})+\cos(\frac{8\pi% }{9}) as

12⁢[e2⁢π9⁢i+e−2⁢π9⁢i+e4⁢π9⁢i+e−4⁢π9⁢i+e6⁢π9⁢i+e−6⁢π9⁢i+e8⁢π9⁢i+e−8⁢π9⁢i]\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⁢π2\pi to anything in the form ea⁢ie^{ai}, this has no effect (again, Euler’s formula and the fact that 2⁢π2\pi radians is a full rotation). Therefore, the previous expression is the same as

12⁢[e2⁢π9⁢i+e16⁢π9⁢i+e4⁢π9⁢i+e14⁢π9⁢i+e6⁢π9⁢i+e12⁢π9⁢i+e8⁢π9⁢i+e10⁢π9⁢i]\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

12⁢[e2⁢π9⁢i+e4⁢π9⁢i+e6⁢π9⁢i+e8⁢π9⁢i+e10⁢π9⁢i+e12⁢π9⁢i+e14⁢π9⁢i+e16⁢π9⁢i]\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 11).

Thus we can write that

12⁢[cos⁡(2⁢π9)+cos⁡(4⁢π9)+cos⁡(6⁢π9)+cos⁡(8⁢π9)]\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] =−1+1+ω+ω2+…+ω8\displaystyle=-1+1+\omega+\omega^{2}+...+\omega^{8}
=−1+1−ω9ω−1\displaystyle=-1+\frac{1-\omega^{9}}{\omega-1}
=−1+1−1ω−1⁢ (ω9=1 as ω is a 9th root of 1)\displaystyle=-1+\frac{1-1}{\omega-1}\text{ ($\omega^{9}=1$ as $\omega$ is a 9% th root of 1)}
=−1\displaystyle=-1

which means that

cos⁡(2⁢π9)+cos⁡(4⁢π9)+cos⁡(6⁢π9)+cos⁡(8⁢π9)=−12\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}