The properties of the cube roots of one: 1 + ω + ω² = 0, ω³ = 1 and the conjugate pair
Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 3
GoalUse 1 + ω + ω² = 0, ω³ = 1 and the conjugate pair to simplify powers and sums of ω.
New words
1 + ω + ω² = 0 · the sum of the three cube roots of unity is zeroω³ = 1 · so ωⁿ only depends on the remainder when n is divided by 3conjugate pair · ω² is the conjugate of ω, and ω × ω² = 1
Explanation
Four facts do almost all the work. (1) The sum of the roots is zero: 1 + ω + ω² = 0. You can see it by adding the algebraic forms, or because ω is a root of Z² + Z + 1 = 0. Rearranged: 1 + ω = −ω², 1 + ω² = −ω and ω + ω² = −1. (2) ω³ = 1, so powers repeat every 3 steps: divide the exponent by 3 and keep the remainder, as in the table. A negative exponent works too, since 1/ω = ω². (3) ω² is the conjugate of ω (the two have the same real part and opposite imaginary parts), they have modulus 1, and ω·ω² = 1, which is the same as 1/ω = ω² and 1/ω² = ω. (4) The difference is imaginary: ω − ω² = √3 i and ω² − ω = −√3 i. A long sum of powers is added in groups of three consecutive terms, each group being 0 by fact (1).
remainder of n ÷ 3
ωⁿ
examples
0
1
ω⁶ = 1, ω³⁰ = 1
1
ω
ω⁷ = ω, ω¹⁰⁰ = ω
2
ω²
ω⁵ = ω², ω⁴⁷ = ω²
Worked examples
Find ω⁴⁷ and ω⁻⁵. 47 = 3 × 15 + 2, so ω⁴⁷ = ω². And ω⁻⁵ = 1/ω⁵ = 1/ω² = ω, because ω⁵ = ω².
Find 1 + ω + ω² + … + ω⁴⁰. There are 41 terms. The first 39 form 13 groups of three, each equal to 0 (ω³ᵐ(1 + ω + ω²)). What is left is ω³⁹ + ω⁴⁰ = 1 + ω = −ω². So the sum is −ω², which is 1/2 + (√3/2)i.
Find (ω − ω²)⁶. Since ω − ω² = √3 i, we get (√3 i)⁶ = 3³·i⁶ = 27 × (−1) = −27.
Class activity
Fill the table. Write ω, ω², ω³, …, ω¹² in a row and mark where the value 1 appears. What rule do you see for the positions of 1, of ω and of ω²?
