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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.

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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=−1−i32=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=ω n mod 3\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=−1−ω\omega^2=-1-\omega.

Also ω⋅ω2=1\omega\cdot\omega^2=1 (they are reciprocals/conjugates), and ∣ω∣=1|\omega|=1.

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.
  1. Reduce exponents mod 3: ω4=ω\omega^4=\omega, ω8=ω2\omega^8=\omega^2.
  2. 1+ω+ω2=01+\omega+\omega^2=0.
Answer:00.
Practice this conceptself-check · 4 quick reps

Try it yourself

If 1,ω,ω21,\omega,\omega^2 are the cube roots of unity, what is (1+ω)(1+ω2)(1+\omega)(1+\omega^2)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    ω3=\omega^3=?
  2. 2.
    1+ω+ω2=1+\omega+\omega^2=?
  3. 3.
    ω2\omega^2 equals which other quantity?
  4. 4.
    1+ω+ω4=1+\omega+\omega^4=?

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]

Drill 2 more on 1, ω, ω² and their identities

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 x2−x+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=ω n mod 3\omega^n=\omega^{\,n\bmod3}, then apply 1+ω+ω2=01+\omega+\omega^2=0.
  • Quadratic roots: x2+x+1=0⇒x=ω,ω2x^2+x+1=0\Rightarrow x=\omega,\omega^2; x2−x+1=0⇒x=−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 n mod 3n\bmod 3 (or 6).

Worked example

If x2+x+1=0x^2+x+1=0, find x2026+x2027x^{2026}+x^{2027}.
  1. Roots are ω,ω2\omega,\omega^2, so take x=ωx=\omega. Reduce: 2026 mod 3=12026\bmod3=1, 2027 mod 3=22027\bmod3=2.
  2. ω2026+ω2027=ω+ω2=−1\omega^{2026}+\omega^{2027}=\omega+\omega^2=-1.
Answer:−1-1.
Practice this conceptself-check · 4 quick reps

Try it yourself

If x2−x+1=0x^2-x+1=0, what is x6x^6?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Roots of x2+x+1=0x^2+x+1=0?
  2. 2.
    Roots of x2−x+1=0x^2-x+1=0?
  3. 3.
    First step on any ωn\omega^n?
  4. 4.
    The three cube roots of kk sum to?

From the bank · past-year question

Example 2Complex NumbersHARD
If x2−x+1=0x^2 - x + 1 = 0, then what is (x−1x)2+(x−1x4)+(x−1x8)\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]

Drill 14 more on applying ω: powers, expressions, related roots

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Complex NumbersEASY
If x,yx,y and zz are the cube roots of unity, then what is the value of xy+yz+zxxy+yz+zx?

[Q4 · Apr · 2024]

Example 2Complex NumbersMODERATE
If ω\omega is a non-real cube root of 1, then what is the value of ∣1−ωω+ω2∣\left|\frac{1-\omega}{\omega+\omega^{2}}\right|?

[Q1 · Apr · 2023]

Example 3Complex NumbersEASY
Which one of the following is correct in respect of the cube roots of unity?

[Q22 · Sep · 2018]

Example 4Complex NumbersMODERATE
What is the value of (−1+i32)3n+(−1−i32)3n\left(\frac{-1+i\sqrt{3}}{2}\right)^{3n} + \left(\frac{-1-i\sqrt{3}}{2}\right)^{3n}, where i=−1i = \sqrt{-1}?

[Q15 · Sep · 2018]

Example 5Complex NumbersMODERATE
The value of (−1+i32)n+(−1−i32)n\left(\dfrac{-1+i\sqrt{3}}{2}\right)^n + \left(\dfrac{-1-i\sqrt{3}}{2}\right)^n, where nn is not a multiple of 3 and i=−1i = \sqrt{-1}, is

[Q6 · Apr · 2017]

Drill every past-year question on this subtopic

18 questions from the bank — paginated, with cart and Word-export support.

Drill the 18 questions

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