Roots of Unity

After reviewing groups from number systems, we now study a finite group that lives inside the complex numbers. The roots of unity are complex numbers whose fixed positive power equals

1

. They form a standard bridge between complex numbers, finite groups, cyclic groups, and geometry on the unit circle. This example is important because it is finite, concrete, and rich enough to show closure, inverses, order of elements, and cyclic generation. In this lesson, students will prove that the set of all

n

-th roots of unity is a finite group under multiplication.

Let

n\in\mathbb{N}

. A complex number

z\in\mathbb{C}

is called an

\textbf{n-th root of unity}

\begin{equation*} z^n=1. \end{equation*}

The set of all

n

-th roots of unity is

\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}

After reviewing groups from number systems, we now study a finite group that lives inside the complex numbers. The roots of unity are complex numbers whose fixed positive power equals

1

n

-th roots of unity is a finite group under multiplication.

Let

n\in\mathbb{N}

. A complex number

z\in\mathbb{C}

is called an

\textbf{n-th root of unity}

\begin{equation*} z^n=1. \end{equation*}

The set of all

n

-th roots of unity is

\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}

The roots of unity are located on the unit circle in the complex plane. They are

\begin{equation*} 1,\; e^{\frac{2\pi i}{n}},\; e^{\frac{4\pi i}{n}},\;\ldots,\; e^{\frac{2\pi i(n-1)}{n}}. \end{equation*}

Equivalently,

\begin{equation*} U_n=\left\{e^{\frac{2\pi ik}{n}}:k=0,1,2,\ldots,n-1\right\}. \end{equation*}

Thus

U_n

has exactly

n

elements.

Let

n\in\mathbb{N}

and let

\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}

Then

(U_n,\cdot)

is a finite group under ordinary complex multiplication.

Given that

n\in\mathbb{N}

and

\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}

To prove that

(U_n,\cdot)

is a finite group. [1] To prove closure. Let

z_1,z_2\in U_n

. Then

\begin{equation*} z_1^n=1,\quad z_2^n=1. \end{equation*}

Since complex multiplication is commutative and associative,

\begin{align*} (z_1z_2)^n &=z_1^n z_2^n\\ &=1\cdot1\\ &=1. \end{align*}

Therefore

z_1z_2\in U_n

. [2] To prove associativity. Let

z_1,z_2,z_3\in U_n

. Since complex multiplication is associative,

\begin{equation*} (z_1z_2)z_3=z_1(z_2z_3). \end{equation*}

[3] To prove the existence of identity element. Since

\begin{equation*} 1^n=1, \end{equation*}

we get

1\in U_n

. Also

\begin{equation*} z\cdot1=1\cdot z=z \quad \forall\, z\in U_n. \end{equation*}

Therefore

1

is the identity element. [4] To prove the existence of inverse elements. Let

z\in U_n

. Then

\begin{equation*} z^n=1. \end{equation*}

Since

z^n=1

, we have

z\neq0

. Also

\begin{align*} z\cdot z^{n-1} &=z^n\\ &=1. \end{align*}

Similarly,

\begin{equation*} z^{n-1}\cdot z=1. \end{equation*}

Thus

z^{n-1}

is the inverse of

z

. Since

\begin{align*} (z^{n-1})^n &=(z^n)^{n-1}\\ &=1, \end{align*}

we get

z^{n-1}\in U_n

. Therefore every element of

U_n

has an inverse in

U_n

. Finally,

\begin{equation*} U_n=\left\{e^{\frac{2\pi ik}{n}}:k=0,1,2,\ldots,n-1\right\}, \end{equation*}

U_n

has

n

elements. Hence,

(U_n,\cdot)

is a finite group.

\square

The proof corrects a subtle point that students sometimes skip. It is not enough to show that

z\cdot z^{n-1}=1

. One must also show that the proposed inverse

z^{n-1}

belongs to the same set

U_n

. This is verified by checking

(z^{n-1})^n=1

. Membership in the original set is part of the inverse axiom.

For

n=4

\begin{equation*} U_4=\{1,i,-1,-i\}. \end{equation*}

Under multiplication, this is a group. The identity element is

1

, the inverse of

i

-i

, and the inverse of

-1

-1

. The element

i

has order

4

, because

\begin{equation*} i^4=1 \end{equation*}

and no smaller positive power of

i

equals

1

For

n=3

\begin{equation*} U_3=\left\{1,\omega,\omega^2\right\}, \end{equation*}

where

\begin{equation*} \omega=e^{\frac{2\pi i}{3}}. \end{equation*}

Then

\begin{equation*} \omega^3=1 \end{equation*}

and

\begin{equation*} 1+\omega+\omega^2=0. \end{equation*}

The group operation is ordinary multiplication, and

\omega

is a generator of the group.

Find the inverse of

e^{\frac{2\pi ik}{n}}

U_n

Let

\begin{equation*} z=e^{\frac{2\pi ik}{n}}\in U_n. \end{equation*}

To find the inverse of

z

U_n

. Since

z^n=1

, we get

\begin{equation*} z^{-1}=z^{n-1}. \end{equation*}

Also,

\begin{align*} z^{-1} &=e^{-\frac{2\pi ik}{n}}\\ &=e^{\frac{2\pi i(n-k)}{n}}. \end{align*}

Therefore the inverse of

e^{\frac{2\pi ik}{n}}

\begin{equation*} e^{-\frac{2\pi ik}{n}}. \end{equation*}

Hence,

\begin{equation*} \left(e^{\frac{2\pi ik}{n}}\right)^{-1}=e^{-\frac{2\pi ik}{n}}. \end{equation*}

\square

Determine the order of

e^{\frac{2\pi i}{6}}

U_6

Let

\begin{equation*} \zeta=e^{\frac{2\pi i}{6}}. \end{equation*}

To determine the order of

\zeta

U_6

. We compute

\begin{equation*} \zeta^6=e^{2\pi i}=1. \end{equation*}

1\leq r<6

, then

\begin{equation*} \zeta^r=e^{\frac{2\pi ir}{6}}\neq1. \end{equation*}

Therefore the least positive integer

r

such that

\zeta^r=1

6

. Hence,

\begin{equation*} o(\zeta)=6. \end{equation*}

\square

The group

U_n

is not just finite; it is cyclic. The element

\begin{equation*} e^{\frac{2\pi i}{n}} \end{equation*}

generates all elements of

U_n

. This fact will become central when cyclic groups are studied in detail. For now, it gives a concrete example where powers of one element produce the entire group.

Use the calculator below to list the exponents of the

n

-th roots of unity. It displays the elements in the form

e^{2\pi ik/n}

and identifies a natural generator. This helps students see that the group has exactly

n

elements.

[1] Define

U_n

. [2] Prove that

U_n

is closed under complex multiplication. [3] Find

U_2

. [4] Find

U_4

. [5] Find the inverse of

e^{\frac{2\pi i\cdot 3}{7}}

U_7

[1]

U_n=\{z\in\mathbb{C}:z^n=1\}

. [2] If

z_1^n=1

and

z_2^n=1

, then

(z_1z_2)^n=z_1^nz_2^n=1

. [3]

U_2=\{1,-1\}

. [4]

U_4=\{1,i,-1,-i\}

. [5] The inverse is

e^{-\frac{2\pi i\cdot 3}{7}}=e^{\frac{2\pi i\cdot 4}{7}}

It is a complex number

z

satisfying

z^n=1

It has exactly

n

elements.

The identity element is

1

It is closed under multiplication, multiplication is associative,

1

is the identity, and every element has an inverse in

U_n

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BMLabs Mathematics

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