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: \(e^{i \theta} = cos ⁡ \theta + i sin ⁡ \theta\) for any real $\theta$. The existence of precisely $n$ solutions follows from the fundamental theorem of algebra: 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 $e^{i n \theta} = 1$. 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:

\(z_{k} = e^{2 \pi i k / n}\) \(k = 0 , 1 , \ldots , n - 1\)

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

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 $z_{n - k} = \overset{―}{z_{k}}$.

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:

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 $z_{k} \rightarrowtail k$.

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 $z_{k} \rightarrowtail k$, 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 $- ( z_{0} + z_{1} + \hdots + z_{n - 1} )$ 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.

totient functioncyclotomic polynomialssymmetryangular spacingregular polygonproduct of rootssum of rootsclosuregroup orderprimitive rootsgeneratorssubgroupscyclic groupangle spacingprincipal rootcomplex rootsunit circleexponential formnth rootspropertiesstructuredefinition