JEE Mains Maths · Complex Numbers

Roots of Unity

The solutions of z to the power n equal to 1 — above all the cube roots 1, ω and ω squared — and the two facts that simplify every power of them.

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Why this matters

Fourteen PYQs, nine of them numerical answers. Ten use the cube roots of unity, usually as the roots of x² + x + 1 = 0, to sum long runs of powers; four use the fifth or sixth roots. Every one reduces a power by its remainder. Two ideas cover the page.

Concept 1 of 2: Cube roots of unity

The roots of

x^3=1

are

1

\omega

and

\omega^2

, with

\omega=\frac{-1+i\sqrt3}2=e^{2\pi i/3}

. Two facts do almost all the work:

\omega^3=1

, so a power of

\omega

depends only on the exponent mod 3; and

1+\omega+\omega^2=0

, so

1+\omega=-\omega^2

and

1+\omega^2=-\omega

. The non-real roots are exactly the roots of

x^2+x+1=0

Definition

$\omega=\frac{-1+i\sqrt3}2$ , $\omega^2=\bar\omega=\frac1\omega$ .
$\omega^3=1$ and $1+\omega+\omega^2=0$ .
$\omega^n+\omega^{-n}$ is 2 when $3\mid n$ and $-1$ otherwise.
$x^2+x+1$ has roots $\omega,\omega^2$ ; $x^2-x+1$ has roots $-\omega,-\omega^2$ .
A real polynomial $P$ is divisible by $x^2+x+1$ exactly when $P(\omega)=0$ .

The two facts

\omega^3=1,\qquad1+\omega+\omega^2=0

Worked example

Simplify

(1+\omega-\omega^2)^3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q67Moderate

Example 1 · Complex Numbers · Roots of Unity

x^{2}+ x + 1 = 0

, then the value of

\left( x +\frac{1}{x} \right)^{4}+\left( x^{2}+\frac{1}{x^{2}} \right)^{4}+\left( x^{3}+\frac{1}{x^{3}} \right)^{4}+ \ldots +\left( x^{25}+\frac{1}{x^{25}} \right)^{4}

is :

Count the multiples of 3

In a sum over

n=1

N

, the terms with

3\mid n

behave differently, and there are

\lfloor N/3\rfloor

of them. Miscounting them by one is the usual slip.

Drill 9 more on cube roots of unity

Concept 2 of 2: The nth roots of unity

The

n

solutions of

z^n=1

are

e^{2\pi ik/n}

k=0,1,\dots,n-1

: equally spaced on the unit circle, so they add to 0. The polynomial

x^{n-1}+\dots+x+1

\frac{x^n-1}{x-1}

, so its roots are the nth roots other than 1, and each satisfies

\alpha^n=1

: powers reduce mod

n

. Other factorisations lead to the same place, for example

x^4+x^2+1=\frac{x^6-1}{x^2-1}

Definition

$z^n=1$ : $z=e^{2\pi ik/n}$ , $k=0,\dots,n-1$ ; their sum is 0.
$x^{n-1}+\dots+x+1=\frac{x^n-1}{x-1}$ .
$z^n=a$ : multiply these by one root of $a$ , $|a|^{1/n}e^{i\arg(a)/n}$ .
The roots form a regular $n$ -gon on the circle.

Roots of unity

z^n=1\ \Rightarrow\ z=e^{2\pi ik/n},\quad k=0,1,\dots,n-1

Worked example

Find the sum of the 100th powers of the roots of

x^3+x^2+x+1=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q62Moderate

Example 2 · Complex Numbers · Roots of Unity

\alpha,\beta,\gamma,\delta

are the roots of the equation

x^{4}+x^{3}+x^{2}+ x + 1 = 0

, then

\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}

is equal to.

The root 1 is missing

x^{n-1}+\dots+1=0

has only the

n-1

roots other than 1. So the sum of their

k

th powers is

n-1

when

n\mid k

, and

-1

otherwise.

Drill 3 more on the nth roots of unity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Cube roots of unity
The two facts
$\omega^3=1,\qquad1+\omega+\omega^2=0$
The nth roots of unity
Roots of unity
$z^n=1\ \Rightarrow\ z=e^{2\pi ik/n},\quad k=0,1,\dots,n-1$

Watch out for (2)

Count the multiples of 3
→ Cube roots of unity
The root 1 is missing
→ The nth roots of unity

Test yourself on Complex Numbers

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Complex Numbers chapter test Drill all 14 questions