In short

The nth roots of unity are the n solutions of z^n = 1. They are z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} for k = 0, 1, \dots, n-1, equally spaced around the unit circle like the vertices of a regular n-gon. For n = 3, the non-real roots are called cube roots of unity and written \omega and \omega^2, where \omega = \frac{-1 + i\sqrt{3}}{2}. A key property: the nth roots of unity always sum to zero.

Take any regular polygon — a triangle, a square, a pentagon — and place it on the complex plane so that one vertex sits at 1 on the real axis and the centre is at the origin. The vertices of that polygon are the nth roots of unity.

This is a striking claim. Why should the solutions of a purely algebraic equation z^n = 1 arrange themselves into a perfectly symmetric geometric figure? The answer comes from De Moivre's theorem and the polar form of complex numbers.

Finding the nth roots of unity

You want all complex numbers z such that z^n = 1. Write z in polar form: z = \cos\theta + i\sin\theta (the modulus must be 1, since |z^n| = |z|^n = 1 forces |z| = 1). By De Moivre's theorem:

z^n = \cos(n\theta) + i\sin(n\theta) = 1

For this to equal 1 = \cos 0 + i\sin 0, you need n\theta = 2k\pi for some integer k. So:

\theta = \frac{2k\pi}{n}

As k runs through 0, 1, 2, \dots, n-1, you get n distinct angles equally spaced around the circle, each separated by \frac{2\pi}{n} radians. At k = n the angle is 2\pi, which is the same point as k = 0, so the cycle repeats.

$n$th roots of unity

The n solutions of z^n = 1 are

z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \qquad k = 0, 1, 2, \dots, n-1.

In exponential notation, z_k = e^{2\pi i k/n}. These n points are equally spaced on the unit circle, forming a regular n-gon with one vertex at 1.

A single symbol captures all the roots. Let \omega = e^{2\pi i/n} = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}. This is the root at k = 1 — the first one after 1 in the counterclockwise direction. Then the n roots are simply:

1, \; \omega, \; \omega^2, \; \omega^3, \; \dots, \; \omega^{n-1}

Each root is the previous one multiplied by \omega, which rotates the point by 2\pi/n around the circle. The root \omega is called a primitive nth root of unity because its successive powers generate all the others.

The sixth roots of unity equally spaced on the unit circleA unit circle on the complex plane with six points equally spaced around it, forming a regular hexagon. The points are labelled 1, omega, omega squared, omega cubed which is negative 1, omega to the fourth, and omega to the fifth. Lines connect adjacent vertices to show the hexagonal shape. The angle between consecutive roots is 60 degrees or pi over 3. ReIm 2π/6 1 ω ω² ω³ = −1 ω⁴ ω⁵ Six roots, equally spaced at 60° apart, forming a regular hexagon.
The sixth roots of unity. The six solutions of $z^6 = 1$ sit at angles $0, \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3$ on the unit circle. They form a regular hexagon. Each root is $\omega$ times the previous one, where $\omega = e^{i\pi/3}$. The root at angle $\pi$ is $-1$, which is indeed a sixth root of unity since $(-1)^6 = 1$.

The cube roots of unity

The case n = 3 is the most heavily tested in exams and the richest in algebraic properties. The three cube roots of unity solve z^3 = 1.

Set \omega = e^{2\pi i/3} = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}. Compute the real and imaginary parts:

\cos\frac{2\pi}{3} = -\frac{1}{2}, \qquad \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}

So:

\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}

The three cube roots of unity are 1, \omega, and \omega^2. What is \omega^2?

\omega^2 = e^{4\pi i/3} = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}

Notice that \omega^2 is the complex conjugate of \omega: \omega^2 = \bar{\omega}. This is because \omega and \omega^2 are symmetric about the real axis — they sit at angles 120° and 240° (= -120°), which are reflections of each other.

The three cube roots of unity forming an equilateral triangleA unit circle on the complex plane with three points marked: 1 on the positive real axis, omega in the second quadrant at angle 120 degrees, and omega squared in the third quadrant at angle 240 degrees. Lines connecting the three points form an equilateral triangle inscribed in the unit circle. The Cartesian coordinates of each root are labelled. ReIm 120° 1 ω = −½ + i√3/2 ω² = −½ − i√3/2 conjugate pair
The cube roots of unity form an equilateral triangle inscribed in the unit circle. The root $1$ sits on the positive real axis. The roots $\omega$ and $\omega^2 = \bar{\omega}$ are a conjugate pair, symmetric about the real axis. The dashed vertical line highlights the conjugate symmetry.

Properties of cube roots of unity

These properties appear constantly in JEE problems. Each one follows from \omega = e^{2\pi i/3}.

Property 1: \omega^3 = 1. By definition — \omega is a cube root of unity.

Property 2: 1 + \omega + \omega^2 = 0. Factor: z^3 - 1 = (z - 1)(z^2 + z + 1). The roots of z^3 - 1 = 0 are 1, \omega, \omega^2. Since \omega and \omega^2 are roots of z^2 + z + 1 = 0, the sum of all three roots is 1 + \omega + \omega^2 = 0 (by Vieta's formulas on z^3 - 1, or by direct substitution into z^2 + z + 1 = 0 and adding 1).

This property is the workhorse: whenever 1 + \omega + \omega^2 appears in a calculation, replace it with 0.

Property 3: \omega^2 = \bar{\omega} and \omega = \bar{\omega^2}. As computed above, \omega and \omega^2 are complex conjugates. This means |\omega| = |\omega^2| = 1 and \omega \cdot \omega^2 = |\omega|^2 = 1 (consistent with \omega^3 = 1).

Property 4: Powers cycle with period 3. \omega^3 = 1, so \omega^4 = \omega, \omega^5 = \omega^2, \omega^6 = 1, and in general \omega^k = \omega^{k \bmod 3}. When simplifying \omega to a large power, just take the exponent modulo 3.

Property 5: \omega^2 is also a primitive cube root of unity. Its powers are \omega^2, \omega^4 = \omega, \omega^6 = 1 — all three cube roots. So \omega^2 generates the same group as \omega. In fact, for any n, e^{2\pi i k/n} is a primitive nth root of unity if and only if \gcd(k, n) = 1.

Summary of the five key properties of cube roots of unityA visual summary showing five boxed properties: omega cubed equals 1, 1 plus omega plus omega squared equals 0, omega squared equals omega bar, powers cycle mod 3, and omega squared is also primitive. Five key properties of ω (cube root of unity) ω³ = 1 1 + ω + ω² = 0 ω² = ω̄ (conjugate) ωᵏ = ωᵏ ᵐᵒᵈ ³ ω² is also a primitive cube root (its powers generate all three roots)
The five properties you need for cube roots of unity. The red-highlighted property — $1 + \omega + \omega^2 = 0$ — is the one that appears most often in simplification problems.

Sum of nth roots of unity

The property 1 + \omega + \omega^2 = 0 for cube roots generalises beautifully. For any n \ge 2:

\sum_{k=0}^{n-1} \omega^k = 1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0

where \omega = e^{2\pi i/n}.

The proof is a one-line application of the geometric series formula. Since \omega \ne 1:

\sum_{k=0}^{n-1} \omega^k = \frac{\omega^n - 1}{\omega - 1} = \frac{1 - 1}{\omega - 1} = 0

Geometrically, this makes perfect sense: the n roots of unity are equally spaced around the unit circle, so their vector sum — the "centre of mass" — is the origin. If you placed equal weights at each vertex of a regular n-gon centred at the origin, the balance point would be at 0.

More generally, for any integer m that is not a multiple of n:

\sum_{k=0}^{n-1} \omega^{mk} = 0

because \omega^m is also a primitive nth root of unity (or at least another root of unity whose powers cycle through all nth roots). And when m is a multiple of n, each \omega^{mk} = (\omega^n)^{m/n \cdot k}... no — more directly: \omega^{mk} = 1 for all k, so the sum is n. This "orthogonality" property is the engine behind the discrete Fourier transform.

Interactive: roots of unity on the unit circle

Drag the red point along the curve below to change the angle \theta. The readouts show \theta, \cos\theta, and \sin\theta. When the point passes through a cube root of unity (\theta = 0, 2\pi/3, 4\pi/3), notice how the coordinates match the values 1, \omega, and \omega^2.

Interactive cube roots of unity explorer on the unit circleA complex plane with a unit circle. A red draggable point moves along the upper semicircle. Readouts show the angle theta, cos theta, and sin theta updating in real time. Three fixed markers indicate the cube roots of unity at angles 0, 2 pi over 3, and 4 pi over 3. ReIm 1−1 i−i 1 (k=0) ω (k=1) ω² (k=2) drag the red point around the circle
Drag the red point along the unit circle. The readouts show $\theta$, $\cos\theta$, $\sin\theta$, and the real and imaginary parts of $z^3$. When the point sits at a cube root of unity, the readouts for $z^3$ show $\operatorname{Re}(z^3) = 1$ and $\operatorname{Im}(z^3) = 0$ — confirming $z^3 = 1$. At all other positions, $z^3 \ne 1$.

Worked examples

Example 1: Simplify $(1 + \omega - \omega^2)^3$ where $\omega$ is a primitive cube root of unity

A pure algebraic manipulation using the properties of cube roots of unity.

Step 1. Use 1 + \omega + \omega^2 = 0 to replace \omega^2.

From 1 + \omega + \omega^2 = 0, you get \omega^2 = -1 - \omega.

Why: this identity lets you eliminate \omega^2 everywhere, reducing all expressions to linear combinations of 1 and \omega.

Step 2. Substitute into 1 + \omega - \omega^2.

1 + \omega - \omega^2 = 1 + \omega - (-1 - \omega) = 1 + \omega + 1 + \omega = 2 + 2\omega

Why: opening the bracket and collecting terms. The expression simplifies to 2(1 + \omega).

Step 3. Simplify 1 + \omega using the same identity.

From 1 + \omega + \omega^2 = 0: 1 + \omega = -\omega^2.

So 2(1 + \omega) = -2\omega^2.

Why: the identity 1 + \omega + \omega^2 = 0 can be rearranged in three ways — use whichever form is most convenient.

Step 4. Cube the result.

(1 + \omega - \omega^2)^3 = (-2\omega^2)^3 = -8\omega^6 = -8(\omega^3)^2 = -8(1)^2 = -8

Why: \omega^6 = (\omega^3)^2 = 1^2 = 1 since \omega^3 = 1. The cube root property collapses the high power instantly.

Result: (1 + \omega - \omega^2)^3 = -8.

The simplification chain from 1 plus omega minus omega squared to negative 8A flow diagram with four stages. First: the original expression 1 plus omega minus omega squared. Arrow leads to 2 plus 2 omega after substituting omega squared equals negative 1 minus omega. Arrow leads to negative 2 omega squared after using 1 plus omega equals negative omega squared. Arrow leads to the cube: negative 8 omega to the sixth equals negative 8. Each arrow is labelled with the property used. 1 + ω − ω² ω² = −1−ω 2 + 2ω 1+ω = −ω² −2ω² cube both sides −8 (−2)³ · (ω²)³ = −8 · ω⁶ = −8 · 1 = −8 The property ω³ = 1 makes the high power collapse.
The simplification in four steps. Each arrow applies one property of $\omega$. The final step uses $\omega^6 = (\omega^3)^2 = 1$ to reach the answer $-8$. The entire computation stays within algebra — no trigonometry needed, even though $\omega$ is defined trigonometrically.

The answer is a real integer, which is typical of symmetric expressions in \omega. Whenever you see \omega in a JEE problem, the strategy is almost always the same: use 1 + \omega + \omega^2 = 0 and \omega^3 = 1 to reduce everything.

Example 2: Show that the fourth roots of unity sum to zero, and find their product

A geometric verification using the explicit coordinates of the roots.

Step 1. List the four roots of z^4 = 1.

With n = 4, the roots are at angles 0, \pi/2, \pi, 3\pi/2:

z_0 = 1, \quad z_1 = i, \quad z_2 = -1, \quad z_3 = -i

Why: the spacing is 2\pi/4 = \pi/2 = 90°, so the roots sit at the four compass points on the unit circle.

Step 2. Compute the sum.

z_0 + z_1 + z_2 + z_3 = 1 + i + (-1) + (-i) = 0

Why: real parts cancel (1 + (-1) = 0), imaginary parts cancel (i + (-i) = 0). This is the general property: nth roots of unity sum to zero for n \ge 2.

Step 3. Compute the product.

z_0 \cdot z_1 \cdot z_2 \cdot z_3 = 1 \cdot i \cdot (-1) \cdot (-i)

In polar form, each root has modulus 1, so the product has modulus 1. The arguments add: 0 + \pi/2 + \pi + 3\pi/2 = 3\pi. Reducing modulo 2\pi: 3\pi \equiv \pi. So the product is e^{i\pi} = -1.

Why: the product of all nth roots of unity is (-1)^{n+1} when n \ge 2. For n = 4: (-1)^5 = -1. (More precisely, the product equals (-1)^{n-1} — both give -1 for n = 4.)

Step 4. Verify using Vieta's formulas.

The polynomial z^4 - 1 = 0 has roots 1, i, -1, -i. The product of all roots equals (-1)^4 \cdot \frac{(-1)}{1} = -1 (constant term divided by leading coefficient, with sign (-1)^n).

\text{Product} = (-1)^4 \cdot \frac{-1}{1} = -1 \;\checkmark

Why: Vieta's formula for the product of roots of z^n + \cdots + c_0 = 0 is (-1)^n \cdot c_0. For z^4 - 1, c_0 = -1 and n = 4.

Result: The four fourth roots of unity sum to 0 and their product is -1.

The four fourth roots of unity forming a square on the unit circleA unit circle on the complex plane with four points at 1, i, negative 1, and negative i forming a square. Vectors from the origin to each root are drawn. The vector sum is shown to be zero by the cancellation of opposite pairs. The product is labelled as negative 1. ReIm 1 i −1 −i Sum = 0 (opposite vertices cancel). Product = −1.
The fourth roots of unity form a square on the unit circle. The roots $1$ and $-1$ cancel, and $i$ and $-i$ cancel, so the sum is $0$. The product of their arguments is $0 + \pi/2 + \pi + 3\pi/2 = 3\pi \equiv \pi \pmod{2\pi}$, so the product is $e^{i\pi} = -1$. This matches the Vieta's formula computation.

The geometric picture makes the sum-to-zero property obvious — opposite vertices of the square are negatives of each other, so they cancel in pairs. For the product, the polar form argument addition gives the answer directly. This is the power of working on the unit circle: sums are visible as balancing vectors, and products are visible as accumulated rotations.

Common confusions

Going deeper

If you can list the nth roots of unity, use the properties \omega^n = 1 and 1 + \omega + \cdots + \omega^{n-1} = 0 in calculations, and handle cube root problems fluently, you have the core. The material below is for readers who want to see the algebraic structure underneath.

Cyclotomic polynomials

The polynomial z^n - 1 factors over the integers. For n = 6:

z^6 - 1 = (z - 1)(z + 1)(z^2 + z + 1)(z^2 - z + 1)

The factor z^2 + z + 1 has roots \omega, \omega^2 (the primitive cube roots). The factor z^2 - z + 1 has roots e^{i\pi/3}, e^{5i\pi/3} (the primitive sixth roots). These irreducible factors are called cyclotomic polynomials — from the Greek for "circle-cutting," because they cut the circle into arcs.

The nth cyclotomic polynomial \Phi_n(z) has degree \phi(n) (Euler's totient function) and its roots are exactly the primitive nth roots of unity. These polynomials are always irreducible over the rationals — a deep result in algebra.

The discrete Fourier transform

The "orthogonality" property of roots of unity — \sum_{k=0}^{n-1} \omega^{mk} = 0 when n \nmid m and = n when n \mid m — is the mathematical heart of the Fourier transform applied to finite sequences. If you have a sequence a_0, a_1, \dots, a_{n-1}, its discrete Fourier transform is A_j = \sum_k a_k \omega^{jk}. The inverse transform uses the conjugate roots \omega^{-jk}. This is the algorithm behind signal processing, image compression, and the fast multiplication of large numbers.

Gauss and the regular 17-gon

In 1796, a 19-year-old mathematician showed that the regular 17-gon can be constructed with compass and straightedge. The proof uses the fact that 17 is a Fermat prime (17 = 2^{2^2} + 1), and the key step is factoring the 16th-degree cyclotomic polynomial \Phi_{17}(z) into quadratic factors using nested square roots. The constructibility of regular polygons is entirely determined by the factorisation of roots of unity — geometry and algebra meeting through the unit circle.

Where this leads next