Evaluating and proving with ω, and forming equations whose roots are expressions in ω
Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 3
GoalEvaluate expressions in ω, prove identities, and form quadratics whose roots are expressions in ω.
New words
identity · an equation that is true for every allowed value; here, a proof that two expressions in ω are equalsum and product of roots · for the quadratic x² − (sum)x + (product) = 0, the roots add to the sum and multiply to the productreduce · replace each power of ω by 1, ω or ω², then use 1 + ω + ω² = 0
Explanation
Every problem with ω is solved by the same four moves. (a) Reduce each power of ω using ω³ = 1. (b) Replace a group 1 + ω + ω² by 0, or ω + ω² by −1. (c) For a fraction, factor ω or ω² out of the top so that it matches the bottom: for example a + bω² = ω²(aω + b), because ω²·aω = a. Then the fraction collapses to ω². (d) To prove that a number is a root of an equation, put x = ω (or ω²), reduce the powers, and check that the result is 0; both ω and ω² give the same answer, since they are conjugates of each other and the equations have real coefficients. To form a quadratic with given roots p and q, calculate p + q and pq using the rules above and write x² − (p + q)x + pq = 0. The answer may be real even though the roots are not.
ƒFormula
1 + ω + ω² = 0, ω³ = 1, 1/ω = ω²
Replace any sum of the three roots by 0 and reduce every power to 1, ω or ω².
Worked examples
Evaluate (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸). Reduce: ω⁴ = ω and ω⁸ = ω², so the product is [(1 + ω)(1 + ω²)]². And (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³ = 0 + 1 = 1. So the value is 1.
Evaluate [(3 + 2ω²)/(3ω + 2) − (1 + 4ω)/(ω² + 4)]⁶. Since ω²(3ω + 2) = 3 + 2ω², the first fraction equals ω². Since ω(ω² + 4) = 1 + 4ω, the second equals ω. The bracket is ω² − ω = −√3 i, and (−√3 i)⁶ = 27 × i⁶ = −27.
Prove that x = ω satisfies x¹⁴ + x⁷ + 1 = 0. Reduce: ω¹⁴ = ω² (14 = 12 + 2) and ω⁷ = ω. So ω² + ω + 1 = 0. ✓ For x = ω²: ω²⁸ = ω and ω¹⁴ = ω², again ω + ω² + 1 = 0. ✓
Form the quadratic whose roots are ω + 2 and ω² + 2. Sum: ω + ω² + 4 = 3. Product: ω³ + 2ω + 2ω² + 4 = 1 − 2 + 4 = 3. The equation is x² − 3x + 3 = 0. (Its roots (3 ± √3 i)/2 are indeed 2 + ω and 2 + ω².)
Class activity
Spot the trap. A friend says ω is a root of x⁹ + x⁶ + 1 = 0 because it looks like x² + x + 1. Evaluate it for x = ω and decide.
Practice
1
Simplify (4 + 3ω + 3ω²)⁷.
4 + 3(ω + ω²) = 4 − 3 = 1, so the value is 1⁷ = 1.
2 + 5ω = ω(2ω² + 5), so the first fraction is ω. 3 + ω² = ω²(3ω + 1), so the second is ω². The bracket is ω − ω² = √3 i and (√3 i)¹⁰ = 3⁵ i¹⁰ = 243 × (−1) = −243.
4
Is ω a root of x⁹ + x⁶ + 1 = 0?
No. ω⁹ = 1 and ω⁶ = 1, so the value is 1 + 1 + 1 = 3, not 0.
5
Write the quadratic equation that has the roots 2ω + 1 and 2ω² + 1.
