Chapter 3 · Complex Numbers and Equations · Concept 3
3-19 Cube Roots of Unity
Lesson 3-17 showed that x³ = 1 has three solutions, not one. Two of them are imaginary — and they turn out to be so useful that one of them gets its own letter.
Guiding question
What is so special about the two answers to x³ = 1 that nobody notices?
Learning outcome. Find the cube roots of unity and use their properties to simplify expressions.
ω, and its two facts
A number that becomes 1 when cubed is called a cube root of unity. There are three of them, and the one that is complex rather than real is written ω — the Greek letter omega.
ω3 = 1
ω2 + ω + 1 = 0
And the order to use them in.
First use ω³ = 1 to bring the power down. Only after that, if ω² and ω are still lying around, use ω² + ω + 1 = 0 to finish.
Doing it the other way round on something like ω²⁰¹⁰ gets you nowhere at all.
Where the second fact comes from.
ω³ = 1, so ω³ − 1 = 0
(ω − 1)(ω² + ω + 1) = 0
And ω is not 1 — it is the imaginary one — so the first bracket cannot be zero. So the second one is. That step is only allowed because ω ≠ 1, which is exactly why the definition insists on the imaginary root.
The three cube roots of 1 sit on a circle at the corners of a triangle. Drag the power and watch it walk round — every three steps it is home.
ω0 = 1
0 divides by 3 exactly, so you are back at the start. That is ω³ = 1 doing its work.
ω is one of the imaginary cube roots of unity. Find ω⁴ + ω⁵.
- 1
Split each power so that an ω³ appears. That is always the first move.
ω4 = ω × ω3, ω5 = ω2 × ω3Take out as many threes as you can. 4 is 3 + 1, and 5 is 3 + 2.
Work with a partner. For each one, say out loud which fact you are about to use and why before you write anything. If you cannot say why, you are probably reaching for the wrong one.
1. ω is one of the imaginary cube roots of unity. Find the value of each.
Find the value of this without expanding the brackets directly:
(1 − ω + ω2)(1 + ω − ω2)
Why does ω get its own letter, when i already has one and everything here could be written as (−1 + √3 i)/2?
Try writing the Challenge out that way and see how far you get.
Using the wrong fact first.
Bring the power down with ω³ = 1 BEFORE reaching for ω² + ω + 1 = 0. On ω²⁰¹⁰ the second fact is no use at all.
Thinking ω could be 1.
1 is a cube root of unity, but ω is the IMAGINARY one. The whole derivation of ω² + ω + 1 = 0 depends on ω ≠ 1.
Leaving 1/ω in the answer.
1/ω is ω², and 1/ω² is ω. Multiply top and bottom by whatever turns the bottom into ω³.
Expanding when you were told not to.
Rearranging ω² + ω + 1 = 0 gives short names — 1 + ω = −ω², 1 + ω² = −ω. Those turn a page of expanding into two lines.
Forgetting ω² is also a cube root of 1.
All three of 1, ω and ω² cube to 1. ω² is the conjugate of ω, and either of the two imaginary ones can be called ω.
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