n-th Roots of Unity: Geometry, Properties, and Group Structure

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N-th root of unity

The n-th roots of unity are exactly the complex solutions ($z \in \mathbb{C}$) of the polynomial equation $z^n = 1$.

In the real number system, the only solution to $x^n = 1$ is: (i) $x = 1$ if $n$ is odd; (ii) $x = \pm 1$ if $n$ is even

However, in the complex domain, there are precisely $n$ distinct solutions to this equation, distributed symmetrically on the unit circle — the set of points in the complex plane at distance 1 from the origin: $\{z : |z| = 1\}$.

Finding the N-th Roots of Unity

1. Express 1 in Exponential Form. Using Euler’s formula ($e^{i\theta} = \cos\theta + i\sin\theta$), we can write: $1 = e^{i \cdot 0} = \cos(0) + i\sin(0)$. Due to the periodicity of sine and cosine (period $2\pi$): $1 = e^{i \cdot 0} = e^{i \cdot 2\pi} = e^{i \cdot 4\pi} = \cdots = e^{i \cdot 2\pi k}$ for any integer $k$.
2. Solve for $z$. $z^n = e^{i \cdot 2\pi k}$. Taking the $n$-th root: $z = e^{\frac{i \cdot 2\pi k}{n}} = e^{\frac{2\pi i k}{n}}$
3. Identify Distinct Roots. As $k$ varies through integers, we get distinct values only for $k = 0, 1, 2, \ldots, n-1$. For $k = n$: $e^{\frac{2\pi i \cdot n}{n}} = e^{2\pi i} = 1 = z_0$ (repeats)
4. The n-th roots of unity are the solution to the equation zⁿ = 1, where z is a complex number. These roots are given by the formula: $z_k = e^{\frac{2πik}{n}} = cos(\frac{2πk}{n}) + isin(\frac{2πk}{n})$ where k = 0, 1, 2, ···, n-1 and:

• $e^{\frac{2πik}{n}}$ represents the exponential form of complex numbers, derived from Euler’s formula e^iθ = cos(θ) + isin(θ).
• k is the index that determines the specific root, $\frac{2πik}{n}$ is the angle from positive real axis.
• These roots are evenly spaced around the unit circle in the complex plane at angles that are multiples of $\frac{2π}{n}$. They form the vertices of a regular n-gon inscribed in the circle.

Primitive Root of Unity

A n-th root of unity is a complex number z satisfying the equation $z^n = 1$. By the Fundamental Theorem of Algebra, there are exactly n distinct roots in the complex plane.

Among these, a primitive root is a generator of the cyclic group of roots. Formally, $\omega$ is a primitive n-th root of unity if the smallest positive integer power that gives 1 is $n$ itself: (i) $\omega^n = 1$; and (ii) $\omega^k \neq 1$ for any $1 \leq k < n$, e.g., n = 4, the 4th roots are $1, i, -1, -i$.

• $z=i$: Powers are $i, -1, -i, 1$. It took 4 steps to get to 1. i is a primitive root.
• $z=-1$: Powers are $-1, 1$. It took 2 steps. -1 is not a primitive root.

In group theory terms, the order of $\omega$ in the multiplicative group $\mathbb{C}^x$ is exactly n.

The principal (or standard) primitive root is defined using Euler’s formula: $\boxed{\omega_n = \zeta_n = e^{\frac{2\pi i}{n}} = \cos(\frac{2\pi i}{n})+i\sin(\frac{2\pi i}{n})}$. This is the root corresponding to $k = 1$ and the complex number on the unit circle with the smallest positive angle $\theta = \frac{2\pi i}{n}$.

All n-th roots of unity can be expressed as powers of the primitive root: $\{ z_0,z_1,\ldots ,z_{n-1}\} =\left\{ e^{\frac{2\pi ik}{n}}:k=0,1,\ldots ,n-1\right\} =\{ 1,\omega _n,\omega _n^2,\ldots ,\omega _n^{n-1}\}$. Each root lies on the unit circle, equally spaced by angle $\frac{2\pi}{n}$.

Primitive Condition. A root $z_k = e^{\frac{2\pi i k}{n}}$ is primitive if and only if $\gcd(k, n) = 1$.

This means $k$ and $n$ must be coprime (they share no common factors other than 1).

Proof.

Part 1: If $\gcd(k, n) = d > 1$, then $z_k$ is NOT primitive.

If gcd(k, n) = d > 1, let $k = d\cdot k'$ and $n = d\cdot n'$. Notice that because $d > 1$, $n' = n/d$ is strictly smaller than $n$ ($n' < n$). Now, let’s raise $z_k$ to the power of this smaller number $n'$:

$z_k^{n'} = (e^{\frac{2\pi i k}{n}})^{n'} = e^{\frac{2\pi i (d \cdot k') n'}{d \cdot n'}} = e^{2\pi ik'} = 1$ (Note: $e^{2\pi i \times \text{integer}}$ is always 1).

Conclusion: We found a power $n' < n$ such that $z_k^{n'} = 1$. Therefore, $z_k$ is not a primitive root.

Part 2: If $\gcd(k, n) = 1$, then $z_k$ is indeed primitive.

Suppose $\gcd(k, n) = 1$. We want to show that $z_k^m \neq 1$ for any $0 < m < n$.

Let’s assume for the sake of contradiction there is some power $m$ such that $z_k^m = 1$. We will analyze what this implies about $m$, $z_k^m = e^{\frac{2\pi i k m}{n}} = 1$.

For a complex exponential $e^{i\theta}$ to equal 1, the exponent $\theta$ must be an integer multiple of $2\pi$, $\frac{2\pi k m}{n} = 2\pi J \quad (\text{where } J \text{ is an integer}) \implies \frac{km}{n} = J \implies$ n divides km. However, we know that $\gcd(k, n) = 1$ ($k$ and $n$ have no common factors). By Euclid’s Lemma, if $n$ divides $km$ and $\gcd(n, k) = 1$, then $n$ must divide $m$.

If $n$ divides $m$, then $m$ must be at least as big as $n$ (since $m > 0$): $m \ge n$. The power $m$ cannot be smaller than $n$. In words, the smallest positive power that results in 1 is $n$ itself. Therefore, $z_k$ is primitive.

$n$	Roots of Unity	Primitive roots	Count $\phi(n)$
2	{1, -1}	-1 (k = 1)	1
3	{1, $\omega, \omega^2$} where $\omega = e^{\frac{2\pi i}{3}}$	$w, w^2$ (k = 1, 2)	2
4	{1, i, -1, -i}	$i, -i$ (k = 1, 3)	2
5	$1, ζ_5, ζ_5^2, ζ_5^3, ζ_5^4$	All except 1 (k = 1, 2, 3, 4)	4

The number of primitive n-th roots of unity is $\phi(n)$, Euler’s totient function.

Examples of N-th Roots of Unity

• Square Roots of Unity (n = 2): z² = 1.

k = 0, $z_0 = e^{\frac{2πi·0}{2}}=e^0=1$. k = 1, $z_1 = e^{\frac{2πi·1}{2}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1.$ The square roots of unity are 1 and −1, lying on opposite ends of the real axis on the unit circle.

• Cube Roots of unity (n = 3)

n = 3, k = 0, $z_0 = e^{\frac{2πi·0}{3}}=e^0=1, z_1 = e^{\frac{2πi}{3}} = cos(\frac{2πi}{3}i)+isin(\frac{2πi}{3}i) = \frac{-1}{2}+i\frac{\sqrt{3}}{2}, z_2 = e^{\frac{4πi}{3}} = \frac{-1}{2}-i\frac{\sqrt{3}}{2}$

Roots: $\left\{1, -\frac{1}{2} + i\frac{\sqrt{3}}{2}, -\frac{1}{2} - i\frac{\sqrt{3}}{2}\right\}$. Often denoted $\{1, \omega, \omega^2\}$ where $\omega = e^{\frac{2\pi i}{3}}$.

• Fourth Roots of Unity (n = 4): z⁴ = 1.

k = 0, $z_0 = e^{\frac{2πi·0}{4}} = e^0 = 1$. k = 1, $z_1 = e^{\frac{2πi·1}{4}} = e^{\frac{πi}{2}} = cos(\frac{π}{2})+isin(\frac{π}{2}) = 0 + i·1 = i$. k = 2, $z_2 = e^{\frac{2πi·2}{4}}=e^{πi} = cos(π)+isin(π) = -1 + i·0 = -1$. k = 3, $z_3 = e^{\frac{2πi·3}{4}} = e^{\frac{3πi}{2}} = cos(\frac{3π}{2})+isin(\frac{3π}{2}) = 0 + i·(-1) = -i$

Properties: $1 + \omega + \omega^2 = 0$ and $\omega^2 = \bar{\omega}$.

The fourth roots of unity are 1, i, −1, and -i, which are positioned at equal angles of ^π⁄₂ radians around the unit circle. They are the vertices of an equilateral triangle inscribed in unit circle.

• Fifth Roots of Unity ($n = 5$): Solving $z^5 = 1$ means finding all complex numbers whose fifth power lands back at 1.

The fifth roots of unity are $\{1, e^{\frac{2\pi i}{5}}, e^{\frac{4\pi i}{5}}, e^{\frac{6\pi i}{5}}, e^{\frac{8\pi i}{5}} \}$. Often denoted as $\{ 1, ζ_5, ζ_5^2, ζ_5^3, ζ_5^4 \}$ where we are using the primitive root $ζ_5 = cos(\frac{2π}{5})+isin(\frac{2π}{5})$. A delightful fact: $\cos\left(\frac{2\pi}{5}\right) = \frac{\sqrt{5} - 1}{4} = \frac{\phi - 1}{2}$ where $\phi = \frac{1 + \sqrt{5}}{2}$ is the golden ratio (the geometry of the regular pentagon is deeply tied to the golden ratio).

Each root corresponds to a rotation by $72^{\circ}$ (since $360^{\circ}/5=72^{\circ}$). They are the vertices of a a regular pentagon inscribed in unit circle.

In the complex plane, the n-th roots of unity are arranged as the vertices of a regular n-sided polygon inscribed in the unit circle. Each vertex corresponds to one of the roots, and the angle between any two consecutive roots is $\frac{2π}{n}$ radians.

Properties of N-th Roots of Unity

1. Distinctness: All $n$ roots $z_0, z_1, \ldots, z_{n-1}$ are distinct. Proof. If $z_j = z_k$ for $0 \leq j < k \leq n-1$, then: $e^{\frac{2\pi i j}{n}} = e^{\frac{2\pi i k}{n}} \implies$
   $e^{\frac{2\pi i (k-j)}{n}} = 1$. This requires $\frac{k-j}{n}$ to be an integer, but $0 < k - j < n$. Contradiction $\blacksquare$

2. Equally Spaced on the Unit Circle: Each root is obtained by rotating the previous one by $\frac{2π}{n}$ radians $z_{k+1} = z_k \cdot e^{\frac{2\pi i}{n}} = z_k \cdot \omega_n$, where $\omega _n=e^{2\pi i/n}$ is a primitive n-th root of unity and represents a rotation by exactly $\frac{2\pi}{n}$ radians, meaning they are evenly (or equally) distributed around the circle.

   Starting from $z_0=1$, each subsequent root is obtained by multiplying by $\omega _n$: $z_{k+1}=z_k\cdot \omega _n$. Unfolding this recursion gives: $z_k=\omega _n^k$, so the entire set is $\{ 1,\; \omega _n,\; \omega _n^2,\; \dots ,\; \omega _n^{n-1}\}$. Multiplying by $\omega _n$ corresponds to rotating a point on the unit circle by: $\frac{2\pi }{n}\mathrm{radians}$. Since this rotation is constant, the points land at angles: $0, \frac{2\pi}{n}, \frac{4\pi}{n}, \dots, \frac{2\pi (n-1)}{n}$, which divide the full circle into n equal arcs.

   Geometrically, the roots form the vertices of a regular n‑gon inscribed in the unit circle.

3. Closure under multiplication. The product of any two n-th roots of unity is also an n-th root of unity: $z_k·z_m = e^{\frac{2πik}{n}}e^{\frac{2πim}{n}} = e^{\frac{2πi(k+m)}{n}} = z_{(k + m)~ mod~ n}$

4. Modulus Equals 1. All roots lie on the unit circle, $|z_k| = \left|e^{\frac{2\pi i k}{n}}\right| = 1$.

5. Inverse of Each Root. Each n-th root of unity has an inverse, which is its complex conjugate and is also an n-th root: $z_k^{-1} = \bar z_k = e^{-\frac{2πik}{n}} = e^{\frac{2\pi i(n - k)}{n}} =​ z_{n-k}$.

6. Sum of All N-th Roots of Unity is zero. The sum of all n-th roots of unity is zero: $\boxed{\sum_{k=0}^{n-1} z_k = \sum_{k=0}^{n-1} \omega_n^k = 0}$.

   Proof: For a primitive n-th root of unity $\omega _n=e^{2\pi i/n}$, the full set of roots is $1, \omega _n, \omega _n^2, \dots, \omega _n^{n-1}$. Since $w_n \ne 1$, the finite geometric series applies: $S_n = \frac{a_1(1-r^n)}{1-r}$ (where the common ratio $r \ne 1, a_1$ is the first term of the series):
   $\sum_{k=0}^{n-1} \omega_n^k = \frac{1 - \omega_n^n}{1 - \omega_n} =[\omega _n^n=e^{2\pi i}=1] \frac{1 - 1}{1 - \omega_n} = 0$

   The n-th roots of unity lie at the vertices of a regular n-gon centered at the origin. Each point is obtained by rotating the previous one by $\frac{2\pi }{n}$ radians. Because the polygon is perfectly symmetric, so their vector sum cancels out.

7. Product of All Roots. $\prod_{k=0}^{n-1} z_k = (-1)^{n+1}$.

8. Sum of Powers. $\sum_{k=0}^{n-1} \omega_n^{jk} = \begin{cases} n & \text{if } n \mid j \\ 0 & \text{if } n \nmid j \end{cases}$. This is the orthogonality relation for roots of unity.

9. Polynomial Factorization. $z^n - 1 = \prod_{k=0}^{n-1} (z - z_k) = \prod_{k=0}^{n-1} \left(z - e^{\frac{2\pi i k}{n}}\right)$

   Proof:

   The root of the polynomial $z^n -1$ are the solution to the equation $z^n = 1$. This equation has exactly the n-th roots of unity as its solutions: $z_k=e^{2\pi ik/n},\quad k=0,1,\dots ,n-1.$

   By the Fundamental Theorem of Algebra, a degree‑n polynomial with these roots must factor as: $z^n - 1 = (z-z_0)(z-z_1) \cdots (z-z_{n-1}) = \prod_{k=0}^{n-1} (z - z_k)$. Replacing $z_k$ with the exponential form $e^{\frac{2\pi ik}{n}}$ results in the final identity.

The Cyclic Group

In modular arithmetic, the integers coprime (relatively prime) to n form a group under multiplication modulo n, denoted $(\mathbb{Z}/n\mathbb{Z})^x$, with order φ(n) (Euler’s totient function), e.g., $(\mathbb{Z}/6\mathbb{Z})^x = \{1, 5 \}$. This is not the same as $\mathbb{Z}/n\mathbb{Z}$ which is the additive group of integers modulo n, e.g., $\mathbb{Z}/n\mathbb{Z} = \{ 0, 1, 2, 3, 4, 5 \}$.

The set of n-th roots of unity $\mu_n = \{ 1,ζ_n,ζ_n^2, \ldots ,ζ_n^{n−1} \}$ where $ζ_n = w_n = e^{\frac{2\pi i}{n}}$ and $\zeta_n^n = 1$ forms a cyclic group of order n under multiplication with $ζ_n$ as a generator.

This group is finite (size n), abelian, cyclic, generated by any primitive n-th root of unity, and isomorphic to the additive group $\mathbb{Z}/n\mathbb{Z}$.

Property	Value
Abelian (cyclic groups are always abelian)	Operation: Multiplication of complex numbers
Identity	$1$ (corresponding to k = 0 in $\mathbb{Z}/n\mathbb{Z}$)
Inverse of $\omega_n^k$	$(\omega _n^k)^{-1}=\omega _n^{-k}=\omega _n^{n-k}$.
Generator	$\omega_n$ (or any primitive n-th root)
Order	$n$ (finite cyclic group)

Isomorphism with $\mathbb{Z}/n\mathbb{Z}$

There exists a natural group isomorphism: $\phi: \mathbb{Z}/n\mathbb{Z} \to \mu_n$, $\phi(k) = \omega_n^k = e^{\frac{2\pi i k}{n}}$. $\mathbb{Z}/n\mathbb{Z}$ corresponds to rotating the unit circle by increments of $\frac{2\pi }{n}$. The isomorphism identifies “addition of angles” in $ \mathbb{Z}/n\mathbb{Z}$ with “multiplication of complex numbers" in $\mu_n$.

• Homomorphism: $\phi(j + k) = \omega_n^{j+k} = \omega_n^j \cdot \omega_n^k = \phi(j) \cdot \phi(k)$
• Surjective: Every n-th root of unity is of the form $\omega_n^k=e^{\frac{2\pi i k}{n}}$ for some k∈{0, 1, …, n−1}, so ϕ hits all of $\mu_n$.
• Injective. If ϕ(j) = ϕ(k), then $e^{\frac{2\pi j k}{n}} = e^{\frac{2\pi i k}{n}}$, which implies j ≡ k (mod n).
• Therefore, $\mu_n \cong \mathbb{Z}/n\mathbb{Z}$. There is an isomorphism between $\mathbb{Z}/n\mathbb{Z}$ and the group of nth roots of unity.

Subgroup Structure: Roots of Unity and Divisors

The subgroups of a cyclic group of order n correspond bijectively to the positive divisors of n. For each divisor d of n, the set $\mu_d = \{ z \in \mathbb{C} | z^d = 1 \}$ is a subgroup of $\mu_n$. Indeed, if d∣n, then the elements of $\mu_d$ are exactly those nth roots of unity whose order divides d.

Example, n = 12; the divisors of 12 are 1, 2, 3, 4, 6, 12. The corresponding subgroups $\mu_d$ of $\mu_{12}$ are: $\mu_1 = \{1\}, \mu_2 = \{1, -1\}, \mu_3 = \{1, \omega_3, \omega_3^2\}, \mu_4 = \{1, i, -1, -i\}, \mu_6 = \{1, \omega_6, \omega_6^2, \omega_6^3, \omega_6^4, \omega_6^5\}, \mu_{12}$ = all 12th roots. Each $\mu_d$ is cyclic of order d, generated by $e^{\frac{2\pi i}{d}}$.