# Roots of Unity

Source: algebrica.org — CC BY-NC 4.0
https://algebrica.org/roots-of-unity/

## Definition

Given a positive integer \\(n\\), a root of unity of order \\(n\\) is a [complex number](../complex-numbers-introduction.md)  \\(z\\) satisfying the [equation](../equations.md) \\[ z^n = 1 \\] 

There are exactly \\(n\\) such numbers in the complex plane, and they can be described with complete explicitness using Euler's formula which states that:
\\[
e^{i\theta} = \cos\theta + i\\,\sin\theta
\\]
for any real \\(\theta\\). The existence of precisely \\(n\\) solutions follows from the [fundamental theorem of algebra](../roots-of-a-polynomial.md): the polynomial \\(z^n - 1\\) has degree \\(n\\) and therefore admits at most \\(n\\) roots in \\(\mathbb{C}\\), and one can verify directly that all \\(n\\) candidates constructed below are distinct.

To derive the explicit form of the roots, one writes a complex number of unit modulus as \\(z = e^{i\theta}\\) and imposes the condition \\(e^{in\theta} = 1\\). Since the complex exponential is periodic with period \\(2\pi\\), this requires \\(n\theta = 2\pi k\\) for some integer \\(k\\), giving \\(\theta = 2\pi k/n\\). As \\(k\\) ranges over any \\(n\\) consecutive [integers](../integers.md), the resulting values of \\(\theta\\) produce \\(n\\) distinct points on the unit circle. It is customary to take \\(k = 0, 1, \ldots, n-1\\), which yields the \\(n\\)-th roots of unity in the form:

\\[z_k = e^{2\pi i k/n}\\]
\\[k = 0, 1, \ldots, n-1\\]

Expanding via Euler's formula, each root can be written in rectangular coordinates as:

\\[
z_k = \cos\\!\left(\frac{2\pi k}{n}\right) + i\\,\sin\\!\left(\frac{2\pi k}{n}\right)
\\]

For \\(k = 0\\) one recovers \\(z_0 = 1\\), which is always a root regardless of \\(n\\). When \\(n = 2\\) the two roots are \\(1\\) and \\(-1\\). When \\(n = 4\\) the four roots are \\(1, i, -1, -i\\), which are familiar from the arithmetic of the Gaussian integers. For general \\(n\\), the roots come in conjugate pairs: since the arguments \\(2\pi k/n\\) and \\(2\pi(n-k)/n\\) sum to \\(2\pi\\), one has \\(z_{n-k} = \overline{z_k}\\).

- - -
## Group structure

The [set](../sets.md) \\(\mu_n\\) of all \\(n\\)-th roots of unity, equipped with the operation of complex multiplication, forms a [group](../groups.md). Closure holds because \\(z_j \cdot z_k = e^{2\pi i(j+k)/n}\\), which is again an \\(n\\)-th root of unity since:

\\[(z_j z_k)^n = z_j^n z_k^n = 1\\]

The identity element is \\(z_0 = 1\\), and the inverse of \\(z_k\\) is \\(z_{n-k}\\), which coincides with the complex conjugate \\(\overline{z_k}\\) since \\(|z_k| = 1\\). More precisely, one has:

\\[
z_j \cdot z_k = z_{(j+k) \bmod n}
\\]

This shows that \\(\mu_n\\) is a [cyclic group](../groups.md) of order \\(n\\), generated by the single element \\(z_1 = e^{2\pi i/n}.\\) Every other root is a power of \\(z_1\\), since \\(z_k = z_1^k\\). This group is isomorphic to \\(\mathbb{Z}/n\mathbb{Z}\\) under addition modulo \\(n\\), with the isomorphism given explicitly by \\(z_k \mapsto k\\).

In particular, \\(\mu_n\\) is abelian, and its subgroup structure mirrors that of \\(\mathbb{Z}/n\mathbb{Z}\\): for each divisor \\(d\\) of \\(n\\), there is a unique subgroup of order \\(d\\), namely \\(\mu_d\\), which embeds naturally in \\(\mu_n\\).

> The isomorphism with \\(\mathbb{Z}/n\mathbb{Z}\\) is given explicitly by \\(z_k \mapsto k\\), and it preserves the group operation in the sense that multiplication of roots corresponds to addition of indices modulo \\(n\\). Since \\(\mathbb{Z}/n\mathbb{Z}\\) is abelian, so is \\(\mu_n\\): the order in which two roots are multiplied is irrelevant, as \\(z_j z_k = z_k z_j\\) follows immediately from  the commutativity of addition among the exponents.

- - -
## Geometric interpretation

In the complex plane, the \\(n\\)-th roots of unity are located at the vertices of a regular \\(n\\)-gon inscribed in the unit circle, with one vertex fixed at the point \\(1\\) on the real axis. The vertices are equally spaced, with an angular separation of \\(2\pi/n\\) radians between any two consecutive roots.

This geometric regularity is a direct consequence of the uniform spacing of the arguments \\(2\pi k/n\\). As \\(k\\) increases by one unit, the corresponding point on the unit circle advances by a fixed angle. The cases \\(n = 3, 4, 6\\) are particularly natural, since the corresponding regular polygons tile the plane. For \\(n = 3\\), for example, one obtains an equilateral triangle, with vertices at:

\begin{align}
z_0 &= 1 \\\\[6pt]
z_1 &= e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}\\,i \\\\[6pt]
z_2 &= e^{4\pi i/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}\\,i
\end{align}

> For \\(n = 6\\) the six roots are the vertices of a regular hexagon, and they include as a subset the roots for \\(n = 2\\) and \\(n = 3\\), which reflects the divisibility \\(2 \mid 6\\) and \\(3 \mid 6\\) and the corresponding subgroup inclusions \\(\mu_2, \mu_3 \subset \mu_6\\).

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## Primitive roots

A root of unity \\(z_k \in \mu_n\\) is called primitive if its order in the group is exactly \\(n\\), meaning that \\(z_k^m \neq 1\\) for every positive integer \\(m < n\\). Equivalently, \\(z_k\\) is a generator of \\(\mu_n\\): every element of the group can be expressed as a power of \\(z_k\\). Since \\(z_k = z_1^k\\), the order of \\(z_k\\) in the cyclic group \\(\mu_n\\) is \\(n / \gcd(k, n)\\). Therefore \\(z_k\\) is primitive if and only if \\(\gcd(k, n) = 1\\).

The number of primitive \\(n\\)-th roots of unity is consequently equal to the number of integers in \\(\{1, 2, \ldots, n\}\\) that are coprime to \\(n\\), which is by definition Euler's totient function \\(\varphi(n)\\).

For instance, when \\(n = 6\\) one has \\(\varphi(6) = 2\\), and the primitive roots are \\(z_1 = e^{\pi i/3}\\) and \\(z_5 = e^{5\pi i/3}\\), corresponding to \\(k = 1\\) and \\(k = 5\\). When \\(n\\) is prime, every root except \\(z_0 = 1\\) is primitive, since \\(\gcd(k, n) = 1\\) for all \\(k \in \{1, \ldots, n-1\}\\), and accordingly \\(\varphi(n) = n - 1\\). If \\(\zeta\\) is any primitive \\(n\\)-th root of unity, then the full set \\(\mu_n\\) can be recovered as:

\\[\{1, \zeta, \zeta^2, \ldots, \zeta^{n-1}\}\\]

This makes the choice of primitive root a matter of convention rather than mathematical substance, since all primitive roots generate the same group.

- - -
## Sum of the roots

The sum of all \\(n\\)-th roots of unity vanishes for every \\(n \geq 2\\). To see this, observe that the [polynomial](../polynomials.md) \\(z^n - 1\\) factors completely over \\(\mathbb{C}\\) as:

\\[
z^n - 1 = (z - z_0)(z - z_1)\cdots(z - z_{n-1})
\\]

Expanding the right-hand side and comparing the coefficient of \\(z^{n-1}\\) on both sides, one finds that this coefficient is zero on the left and equal to \\(-(z_0 + z_1 + \cdots + z_{n-1})\\) on the right. It follows that:

\\[
\sum_{k=0}^{n-1} z_k = 0
\\]

An alternative verification uses the formula for a [geometric series](../geometric-series.md): since \\(z_1 \neq 1\\) when \\(n \geq 2\\), one has:

\\[
\sum_{k=0}^{n-1} z_1^k = \frac{z_1^n - 1}{z_1 - 1} = \frac{1 - 1}{z_1 - 1} = 0
\\]

Geometrically, this result states that the centroid of the vertices of a regular \\(n\\)-gon inscribed in the unit circle coincides with the origin, which is geometrically evident by symmetry.

- - -
## Product of the Roots

The product of all \\(n\\)-th roots of unity is given by the following identity. Since the
constant term of \\(z^n - 1\\) is \\(-1\\) and the leading coefficient is \\(1\\), comparing
coefficients in the factorisation:

\\[
z^n - 1 = (z - z_0)(z - z_1)\cdots(z - z_{n-1})
\\]

yields:

\\[
\prod_{k=0}^{n-1} z_k = (-1)^{n+1}
\\]

This result is a direct consequence of [Vieta's formulas](../trinomials.md), which relate the coefficients of a polynomial to the elementary symmetric polynomials of its roots. For instance, when \\(n = 2\\) the roots are \\(1\\) and \\(-1\\), whose product is \\(-1 = (-1)^3\\), and when \\(n = 3\\) the roots are the three cube roots of unity, whose product is \\(1 = (-1)^4\\).

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## Cyclotomic polynomials

The primitive \\(n\\)-th roots of unity are precisely the roots of the \\(n\\)-th cyclotomic polynomial \\(\Phi_n(x)\\), defined as the monic polynomial whose roots are exactly the primitive \\(n\\)-th roots of unity. We have:

\\[
\Phi_n(x) = \prod_{\substack{k=1 \\ \gcd(k,\,n)=1}}^{n} \left(x - e^{2\pi i k/n}\right)
\\]

The degree of \\(\Phi_n(x)\\) is \\(\varphi(n)\\). The first few examples are as follows: \\(\Phi_1(x) = x - 1\\), \\(\Phi_2(x) = x + 1\\), \\(\Phi_3(x) = x^2 + x + 1\\), and \\(\Phi_4(x) = x^2 + 1\\). An important identity connects the cyclotomic polynomials to the [factorisation](../factoring-ac-method.md) of \\(z^n - 1\\): since every \\(n\\)-th root of unity is a primitive \\(d\\)-th root for exactly one divisor \\(d\\) of \\(n\\), one has:

\\[
z^n - 1 = \prod_{d \mid n} \Phi_d(z)
\\]

This identity allows one to compute cyclotomic polynomials recursively. A fundamental theorem in algebraic number theory asserts that \\(\Phi_n(x)\\) is irreducible over \\(\mathbb{Q}\\) for every positive integer \\(n\\): this is treated in detail in the dedicated entry on cyclotomic polynomials.