NDA Maths · Complex Numbers

Cube Roots of Unity

The three cube roots of 1 — namely 1, ω, ω² — and the two identities (ω³ = 1 and 1 + ω + ω² = 0) that answer a large, predictable family of NDA questions.

Why this matters

Cube roots of unity are the single most reliable pattern in this chapter — and one of the most HARD-concentrated. Almost every question reduces to 'powers of ω cycle every 3' plus '1 + ω + ω² = 0'.

Concept 1 of 2

1, ω, ω² and their identities

Intuition

The equation x3=1x^3=1 has three roots: 11 and the two non-real ones, ω\omega and ω2\omega^2, which are complex conjugates sitting on the unit circle 120° apart. Two facts do all the work: their powers cycle every 3, and the three sum to zero.

Definition

ω=1+i32,  ω2=1i32=ωˉ\omega=\dfrac{-1+i\sqrt3}{2},\;\omega^2=\dfrac{-1-i\sqrt3}{2}=\bar\omega. The two identities:

  • **ω3=1\omega^3=1** — so ωn=ωnmod3\omega^n=\omega^{\,n\bmod 3} (powers cycle every 3).
  • **1+ω+ω2=01+\omega+\omega^2=0** — so ω+ω2=1\omega+\omega^2=-1 and ω2=1ω\omega^2=-1-\omega.

Also ωω2=1\omega\cdot\omega^2=1 (they are reciprocals/conjugates), and ω=1|\omega|=1.

Cube roots of unity identities

ω3=11+ω+ω2=0ωˉ=ω2ωn=ωnmod3\omega^3=1 \qquad 1+\omega+\omega^2=0 \qquad \bar\omega=\omega^2 \qquad \omega^n=\omega^{\,n\bmod 3}
1ωω²120°1 + ω + ω² = 0

Worked example

Simplify 1+ω4+ω81+\omega^4+\omega^8, where ω\omega is a non-real cube root of unity.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Complex NumbersEASY
If 1,ω,ω21, \omega, \omega^2 are the cube roots of unity, then (1+ω)(1+ω2)(1+ω3)(1+ω+ω2)(1+\omega)(1+\omega^2)(1+\omega^3)(1+\omega+\omega^2) is equal to

[Q9 · Apr · 2017]

ω2=ωˉ\omega^2=\bar\omega, but ω2ω\omega^2\ne-\omega

ω2\omega^2 is the conjugate ωˉ\bar\omega (both unit-circle cube roots, 120° apart). From 1+ω+ω2=01+\omega+\omega^2=0 you get ω2=1ω\omega^2=-1-\omeganot ω-\omega. Treating ω2\omega^2 as ω-\omega (or forgetting to reduce ωn\omega^n by nmod3n\bmod 3 first) wrecks the algebra. Also note ωω2=ω3=1\omega\cdot\omega^2=\omega^3=1.

Concept 2 of 2

Applying ω: powers, expressions, related roots

Intuition

Most ω questions are recognition: spot that a given number IS ω (e.g. (1+3)/2(-1+\sqrt{-3})/2), reduce every exponent mod 3, and collapse using 1+ω+ω2=01+\omega+\omega^2=0. The roots of x2+x+1=0x^2+x+1=0 are ω,ω2\omega,\omega^2; the roots of x2x+1=0x^2-x+1=0 are ω,ω2-\omega,-\omega^2; and x3=kx^3=k has roots k1/3{1,ω,ω2}k^{1/3}\{1,\omega,\omega^2\}.

Definition

  • Reduce then collapse: ωn=ωnmod3\omega^n=\omega^{\,n\bmod3}, then apply 1+ω+ω2=01+\omega+\omega^2=0.
  • Quadratic roots: x2+x+1=0x=ω,ω2x^2+x+1=0\Rightarrow x=\omega,\omega^2; x2x+1=0x=ω,ω2x^2-x+1=0\Rightarrow x=-\omega,-\omega^2 (primitive 6th roots).
  • **Cube roots of kk:** the roots of z3=kz^3=k are k1/3,k1/3ω,k1/3ω2k^{1/3},\,k^{1/3}\omega,\,k^{1/3}\omega^2 — they sum to 0 and form an equilateral triangle.
  • Sums like αn+βn\alpha^n+\beta^n for α,β\alpha,\beta cube/6th-roots are a small-case match on nmod3n\bmod 3 (or 6).

Worked example

If x2+x+1=0x^2+x+1=0, find x2026+x2027x^{2026}+x^{2027}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Complex NumbersHARD
If x2x+1=0x^2 - x + 1 = 0, then what is (x1x)2+(x1x4)+(x1x8)\left(x - \frac{1}{x}\right)^2 + \left(x - \frac{1}{x^4}\right) + \left(x - \frac{1}{x^8}\right) equal to?

[Q14 · Apr · 2025]

Summary — formulas & gotchas at a glance

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Formulas (1)

  • 1, ω, ω² and their identities

    Cube roots of unity identities

    ω3=11+ω+ω2=0ωˉ=ω2ωn=ωnmod3\omega^3=1 \qquad 1+\omega+\omega^2=0 \qquad \bar\omega=\omega^2 \qquad \omega^n=\omega^{\,n\bmod 3}

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Drill every past-year question on this subtopic

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