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 $z$ satisfying the equation \[ z^n = 1 \]
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:
for any real $\theta$. The existence of precisely $n$ solutions follows from the fundamental theorem of algebra: 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, 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:
Expanding via Euler's formula, each root can be written in rectangular coordinates as:
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 $\mu_n$ of all $n$-th roots of unity, equipped with the operation of complex multiplication, forms a group. 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:
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:
This shows that $\mu_n$ is a cyclic group 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:
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 $\mu_2, \mu_3 \subset \mu_6$.
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:
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 $z^n - 1$ factors completely over $\mathbb{C}$ as:
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:
An alternative verification uses the formula for a geometric series: since $z_1 \neq 1$ when $n \geq 2$, one has:
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:
yields:
This result is a direct consequence of Vieta's formulas, 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 $1 = (-1)^4$.
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:
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 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:
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.