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Principle deep dive

Cube roots of unity (1 + ω + ω² = 0, ω³ = 1)

Pairs with Vieta in the ω-Vieta compound. Beyond the named subtopic, ω appears explicitly inside Complex Numbers' modulus problems, M&D's special determinants, and Quadratic Equation questions where x² + x + 1 = 0 unlocks ω³ = 1 simplifications.

questions in the bank
29
tagged HARD
41%
chapter spread
3
worked examples below
4

When to reach for it

ω appears, or any expression that can be re-cast as 'cube roots of unity'.

Why this principle matters

Three identities define the toolkit: 1 + ω + ω² = 0, ω³ = 1, and |ω| = 1. Combined, they reduce almost every cube-roots-of-unity question to small-case arithmetic — even when the question hides them behind powers like ω^100 or expressions like (1 + ω − ω²)^100.

The cycle is the secret: ω, ω², 1, ω, ω², 1, ... — every third power returns to 1. So any high power of ω reduces mod 3. ω^100 = ω^(99+1) = ω^1 = ω. The pattern means NDA can write any power and you compute in one step.

Watch for the disguised form: x² − x + 1 = 0 has roots ω, ω² (NOT 1 ± ω). x² + x + 1 = 0 also has cube-root-of-unity roots. Recognising 'this is a cube-roots-of-unity quadratic' is half the work; the rest is applying the three identities.

4 worked examples from the bank

Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.

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 NumbersMODERATE
Consider the following statements in respect of the roots of the equation x3−8=0x^{3}-8=0: 1. The roots are non-collinear. 2. The roots lie on a circle of unit radius. Which of the above statements is/are correct?

[Q34 · Sep · 2021]

Example 4Complex NumbersHARD
If ω≠1\omega \neq 1 is a cube root of unity, then what is (1+ω−ω2)100+(1−ω+ω2)100\left(1+\omega-\omega^{2}\right)^{100}+\left(1-\omega+\omega^{2}\right)^{100} equal to?

[Q4 · Sep · 2024]

Variants to recognise

Same principle, different surfaces. Pattern-match these on test day.

  • 1 + ω + ω² = 0

    The single most useful identity. Lets you rewrite 1 + ω = −ω², ω + ω² = −1, 1 + ω² = −ω.

  • ω³ = 1 (cycling)

    Powers cycle mod 3: ω^(3k) = 1, ω^(3k+1) = ω, ω^(3k+2) = ω². Any high power reduces in one step.

  • ω, ω² as roots of x² + x + 1 = 0

    Vieta gives ω + ω² = −1 and ω · ω² = 1. Bridges directly to the Vieta principle.

  • Geometric placement on the unit circle

    The three cube roots of unity are equally spaced at angles 0, 2π/3, 4π/3 on |z| = 1. Forms an equilateral triangle.

Drill every cube roots of unity (1 + ω + ω² = 0, ω³ = 1) question

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

Drill the 29 questions

Related principles

Often combined with this one — drill these next if you found the examples above tractable.