# Roots of Unity

A **root of unity** is a complex number that, when raised to a positive integer power, results in \(1\). Roots of unity have connections to many areas of mathematics, including the geometry of regular polygons, group theory, and number theory.

The following problem, although not seemingly related to complex numbers, is a good demonstration of how roots of unity work:

Brilli the ant stands on vertex 1 of the regular decagon below.

- He starts by hopping 1 space at a time (from 1 to 2, then from 2 to 3, and so on). He performs 10 hops in this way.
- He then hops 2 spaces at a time (from 1 to 3, then from 3 to 5, and so on). He performs 10 hops in this way.
- He continues to increase the hop distance every 10 hops: hopping 3 spaces 10 times, then hopping 4 spaces 10 times, and so on.
- After Brilli has hopped 10 spaces, 10 times, he ends his workout.

When Brilli has completed his workout, which vertex will he be standing on?

The "hops" of different distances in the previous problem are analogous to the 10\(^\text{th}\) roots of unity. In fact, the 10\(^\text{th}\) roots of unity, when graphed in the complex plane, form a regular decagon. Peforming the 10 "hops" around the decagon is just like raising a 10\(^\text{th}\) root of unity to the 10\(^\text{th}\) power. In the case of Brilli, it brings him back where he started from. In the case of 10\(^\text{th}\) roots of unity, it results in \(1\).

#### Contents

- Definition of Roots of Unity
- Solving for \(\textit{n}^\text{th}\) Roots of Unity
- Solving Equations of the Form \(x^n=a\)
- Relation of Roots of Unity to Regular Polygons
- Rotations using Roots of Unity
- Relation of Roots of Unity to Geometric Progressions
- Properties of Roots of Unity
- Additional Problem Solving with Roots of Unity

## Definition of Roots of Unity

For any positive integer \(n\), the \(\textit{n}^\textbf{th}\)

roots of unityare the complex solutions to the equation \(x^n=1\), and there are will be \(n\) solutions to the equation.

If \(n\) is even, there will be 2 real solutions to the equation \(x^n=1\), which are \(1\) and \(-1\); if \(n\) is odd, there will be 1 real solution, which is \(1\).

What are the 3\(^\text{rd}\) roots of unity?

By definition, the 3\(^\text{rd}\) roots of unity are the solutions to the equation \(x^3=1\).

A keen eye would recognize that \(x=1\) is one of the solutions to this equation, and so \(1\) is a 3\(^\text{rd}\) root of unity.

However, there are two more

complexsolutions to this equation that need to be accounted for. These solutions are \(x=\large{\frac{-1+i\sqrt{3}}{2}}\) and \(x=\large{\frac{-1-i\sqrt{3}}{2}}\).Written as a set, the 3\(^\text{rd}\) roots of unity are \[U_3=\left\{1,\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}\right\}.\ _\square\]

These solutions can be confirmed using complex number arithmetic. You can try this in the example below.

## Solving for \(\textit{n}^\text{th}\) Roots of Unity

Euler's formula can be used to find the \(n^\text{th}\) roots of unity for any positive integer \(n\).

\(e^{ix}=cis(x)=cos(x)+isin(x)\)

Let \(n\) be a positive integer and \(U_n\) be the set of all \(n^\text{th}\) roots of unity. Then

\[U_n=\left\{e^{2k\pi i/n}\mid k\in \{1,2, \ldots , n\}\right\}.\]

It is important to note that set of all \(n^\text{th}\) roots of unity always contains \(n\) elements. Each of these roots of unity can be found by changing the value of \(k\) in the expression \(e^{2k\pi i/n}\).

By Euler's formula,

\[e^{2\pi i}=\cos(2\pi)+i\sin(2\pi)=1.\]

Let \(k\) be any positive integer. Then

\[\big(e^{2\pi i}\big)^k=e^{2k\pi i}=1^k=1.\]

It is given that \(x^n=1\), with \(x\) complex and \(n\) a positive integer. Now we have

\[x^n=1=e^{2\pi i}=e^{4\pi i}=e^{6\pi i}=\ldots=e^{2k\pi i}.\]

Raising each power to the \(\large\frac{1}{n}\) power gives

\[\left(x^n\right)^{1/n}=x=e^{2\pi i/n}=e^{4\pi i/n}=e^{6\pi i/n}=\ldots=e^{2k\pi i/n}.\]

Note that if \(k>n\), then the angle \(\large\frac{2k\pi}{n}\) is co-terminal with \(\large\frac{2(k-n)\pi}{n}\).

Therefore, there are \(n\) distinct solutions to \(x^n=1\) each given by \(x=e^{2k\pi i/n}\) with \(k=1,2,3,\ldots,n\). \(_\square\)

Solve the equation \(x^3=1\) for all \(x.\)

By the theorem above, the solutions are given by

\[\begin{array} &x=e^{2i\pi/3}, &e^{4i\pi/3}, &e^{6i\pi/3} \end{array}.\]

By Euler's formula, these are, respectively,

\[\begin{align} x&=\cos{\left(\frac{2\pi}{3}\right)}+i\sin{\left(\frac{2\pi}{3}\right)}=\frac{-1}{2}+i\frac{\sqrt{3}}{2}\\ x&=\cos{\left(\frac{4\pi}{3}\right)}+i\sin{\left(\frac{4\pi}{3}\right)}=\frac{-1}{2}-i\frac{\sqrt{3}}{2}\\ x&=\cos{\left(\frac{6\pi}{3}\right)}+i\sin{\left(\frac{6\pi}{3}\right)}=1.\ _\square \end{align}\]

## Solving Equations of the Form \(x^n=a\)

Roots of unity can be used to solve any equation of the form \(x^n=a\), where \(a\) is a real number.

## Find all complex solutions to the equation \(x^4=2\).

The method for solving this equation is very similar to solving \(x^4=1\):

\[x^4=2(1)=2e^{2i\pi}=2e^{4i\pi}=2e^{6i\pi}=2e^{8i\pi}.\]

Taking the \(4^\text{th}\) root of each gives

\[\begin{align} x&=\sqrt[4]{2}e^{2k\pi i/4}\ \text{ for }\ k=1,2,3,4\\ \\ x&=\sqrt[4]{2}i, ~-\sqrt[4]{2}i, ~-\sqrt[4]{2}, ~~\sqrt[4]{2}.\ _\square \end{align}\]

## Relation of Roots of Unity to Regular Polygons

Roots of unity have a strong relation to geometry. By graphing the roots of unity on the complex plane, they can be used to generate the vertices of a regular polygon.

For any integer \(n\ge 3\), the \(n^\text{th}\) roots of unity, when graphed on the complex plane, are the vertices of a regular \(n\)-gon.

An equilateral triangle has its centroid located at the origin and a vertex at \((1,0)\). What are the coordinates of the other two vertices?

The point \((1,0)\) corresponds to the complex number \(1+0i\) on the complex plane. This number is a 3\(^\text{rd}\) root of unity. The other 3\(^\text{rd}\) roots of unity will be the remaining vertices of the equilateral triangle on the complex plane:

\[\frac{-1+i\sqrt{3}}{2}\text{ and }\frac{-1-i\sqrt{3}}{2}\]

These complex numbers correspond to the following points on the coordinate plane:

\[\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\text{ and }\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right).\ _\square\]

## Rotations using Roots of Unity

The \(8^\text{th}\) roots of unity and the \(12^\text{th}\) roots of unity together comprise all of the special angles on the complex unit circle.

Because these values can be easily calculated or memorized, they are very useful for performing rotations in the complex plane. By extension, they can be used to perform rotations in any two-dimensional or even three-dimensional space.

The point \((2,3)\) is rotated \(\frac{3\pi}{4}\) radians counterclockwise about the origin. What is the resulting image?

Note that the corresponding point in the complex plane is \(2+3i\). Also note that \(e^{3\pi i/4}\) is an \(8^\text{th}\) root of unity, and its value is \(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\).

The rotation in the complex plane can be achieved by multiplying the complex numbers. This resulting image in the complex plane is

\[\left(\vphantom{\frac{\sqrt{2}}{2}} 2+3i\right)\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=-\sqrt{2}-\frac{3\sqrt{2}}{2}+i\left(\sqrt{2}-\frac{3\sqrt{2}}{2}\right)=-\frac{5\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}.\]

Then the corresponding image in the coordinate plane is \(\left(-\frac{5\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right).\ _\square\)

## Relation of Roots of Unity to Geometric Progressions

From geometric progressions we have

\[\sum_{k=0}^n{x^k}=\frac{x^{n+1}-1}{x-1}.\]

This identity, along with the properties of roots of unity, can be used to find the solutions of certain polynomial equations.

Find all complex solutions of the equation \(x^3+x^2+x+1=0\).

By the identity above, the equation becomes

\[\begin{array} &\dfrac{x^4-1}{x-1}=0 &\text{ or simply } &x^4=1, x\neq 1. \end{array}\]

The solutions will be the 4\(^\text{th}\) roots of unity (with the exception of \(x=1\)):

\[\begin{array} &x=e^{2\pi i/4}, &e^{4\pi i/4}, &e^{6\pi i/4}, &e^{8\pi i/4}. \end{array}\]

Expanding using Euler's formula gives

\[\begin{array}{rrrrr} x=&0+i, &0-i, &-1+0i, &1+0i. \end{array}\]

The last solution is rejected because \(x\neq 1\). Thus, the solutions are \(i, -i,\) and \(-1\). \(_\square\)

## Properties of Roots of Unity

Roots of unity have many special properties and applications. These are just some of them:

- If \(x\) is an \(n^\text{th}\) root of unity, then so is \(x^k,\) where \(k\) is any integer.
- If \(x\) is an \(n^\text{th}\) root of unity, then \(x^n=1\).
- The sum of all \(n^\text{th}\) roots of unity is always zero for \(n\ne 1\).
- The product of all \(n^\text{th}\) roots of unity is always \((-1)^{n+1}\).
- \(1\) and \(-1\) are the only real roots of unity.
- If a number is a root of unity, then so is its complex conjugate.
- The sum of all the \(k^\text{th}\) power of the \(n^\text{th}\) roots of unity is \(0\) for all integers \(k\) such that \(k\) is not divisible by \(n.\)
- The sum of the absolute values of all the \(n^\text{th}\) roots of unity is \(n.\)
- If \(x\) is an \(n^\text{th}\) root of unity not equal to \(1,\) then \(\sum\limits_{k=0}^{n-1}{x^k}=0\).

If \(x\) is an \(n^\text{th}\) root of unity, then so is \(x^k,\) where \(k\) is any integer.

By using index rule, we can deduce that \((x^k)^n=(x^n)^k\). As \(x\) is an \(n^\text{th}\) root of unity, \(x^n=1\), therefore \((x^n)^k=1\), and \((x^k)^n=1\). Hence \(x^k\) is also an \(n^\text{th}\) root of unity.

Find the product of all the \(2016^\text{th}\) roots of unity.

By definition, the product of the roots of unity is the same as the product of the roots of the equation

\[x^{2016}-1=0.\]

By Vieta's formula, the product of roots is related to the constant term of the polynomial. The degree of the polynomial is even, so the product of roots is the same as the constant term, \(-1\). \(_\square\)

Find the sum of all \(1729^\text{th}\) roots of unity.

We are looking for the sum of roots of

\[x^{1729}-1=0.\]

By Vieta's formula, the sum of the roots is the opposite of the coefficient of the first degree term. The coefficient of the first degree term is \(0\), so the sum of roots is \(0\). \(_\square\)

Find the sum of all the \(1000^\text{th}\) powers of the \(17^\text{th}\) roots of unity.

The equation with roots is

\[\begin{align} x^{17}-1&=0\\ x^{17}&=1\\ \left(x^{17}\right)^{58}&=1^{58}\\ &=1\\ x^{986}&=1\\ x^{1000}&=x^{14}. \end{align}\]

So we are just searching for the sum of all the fourteenth power. By Newton's sum (also read this special case) we see that \(s_1=s_2=s_3=s_4=...=s_{14}=0\). Hence we conclude the sum is \(0\). \(_\square\)

This will be true for any \(k^\text{th}\) power such that \(k\) is an integer and is not divisible by \(n.\) \((\)taking the \(k^\text{th}\) power sum of the \(n^\text{th}\) roots of unity\()\)

Find the sum of the absolute values of the roots of \[\] \[x^{172920152016}=1.\]

If it was just the sum, then we would get zero. But the question asks us about modulus. So we use the fact that each root can be expressed as \(e^{ki}\), where \(k\) is a real number. What is \(|e^{ki}|\)? Recalling Euler's identity, we can rewrite it as

\[\left|e^{ki}\right|=\left|\cos k+i\sin k\right|=\sqrt{\cos^2 k+\sin^2 k}=\sqrt{1}=1.\]

So each root has an absolute value of \(1.\) Hence, the sum of the absolute values of the \(172920152016\) roots is \(172920152016\). \(_\square\)

You can generalize for the \(n^\text{th}\) roots of unity.

## Additional Problem Solving with Roots of Unity

If \(\alpha\) is one of the non-real seventh roots of unity, then find the discriminant of the monic quadratic equation with the roots \(\alpha+\alpha^2+\alpha^4\) and \(\alpha^3+\alpha^5+\alpha^6\).

\(\)

**Details and Assumptions:**

- The discriminant of a quadratic equation \(ax^2+bx+c=0\) is \(b^2-4ac.\)

Given \[f(x) = x^{13} + 2x^{12} + 3x^{11} + 4x^{10} + \cdots + 13x + 14,\] denote \[ N = f(a) \times f\left(a^2\right) \times f\left(a^3\right) \times\cdots \times f\left(a^{14} \right),\] where \(a = \cos\left( \dfrac{2\pi}{15} \right) + i \sin\left( \dfrac{2\pi}{15} \right) .\) Then what is the value of \(M \) such that \( N^\frac{1}{M} = 15 ?\)

\[\begin{cases} \alpha^3 \beta^5 = 1 \\ \alpha^7 \beta^2 = 1 \end{cases}\]

Let \(\alpha = \cos \theta_1 + i \sin \theta_1\) and \(\beta = \cos \theta_2 + i \sin \theta_2\) be the complex numbers satisfying the system above, where \(0 < \theta_1\) and \(\theta_2 < \frac{\pi}{2}\).

If \(\frac{\theta_1}{\theta_2} = \frac{a}{b}\), where \(a, b\) are coprime positive integers, compute \(a+b\).