The cube roots of unity: ω, its properties and its uses
Answers are for parents and teachers.
1
Evaluate (1 + 2ω + 3ω²)(1 + 3ω + 2ω²).
Expand and reduce: 1 + 3ω + 2ω² + 2ω + 6ω² + 4ω³ + 3ω² + 9ω³ + 6ω⁴ = 14 + 11ω + 11ω² (using ω³ = 1, ω⁴ = ω). Then 14 + 11(ω + ω²) = 14 − 11 = 3.
2
Find ω⁻¹⁰ + ω⁻²⁰.
ω⁻¹⁰ = 1/ω¹⁰ = 1/ω = ω² and ω⁻²⁰ = 1/ω²⁰ = 1/ω² = ω. The sum is ω + ω² = −1.
3
Prove [(7 + 3ω)/(7ω² + 3) − (5 + 2ω²)/(5ω + 2)]⁴ = 9.
7 + 3ω = ω(7ω² + 3), so the first fraction is ω. 5 + 2ω² = ω²(5ω + 2), so the second is ω². The bracket is ω − ω² = √3 i and (√3 i)⁴ = 9.
4
Prove that x = ω² is a root of x²⁰ + x¹⁰ + 1 = 0.
(ω²)²⁰ = ω⁴⁰ = ω (40 = 39 + 1) and (ω²)¹⁰ = ω²⁰ = ω² (20 = 18 + 2). The sum is ω + ω² + 1 = 0.
5
Write the quadratic equation that has the roots 3 + ω and 3 + ω².
Sum: 6 + (ω + ω²) = 5. Product: 9 + 3(ω + ω²) + ω³ = 9 − 3 + 1 = 7. The equation is x² − 5x + 7 = 0.