Complex Numbers
Cube Roots of Unity
Grade 11
Question:
<p>If <span class="math">\alpha = \cos \frac{2\pi}{7} + i\sin \frac{2\pi}{7}</span>, then the quadratic equation whose roots are <span class="math">\beta = \alpha + \alpha^2 + \alpha^4</span> and <span class="math">\gamma = \alpha^3 + \alpha^5 + \alpha^6</span> is</p>
<p>(a) <span class="math">x^2 - x + 2 = 0</span></p>
<p>(b) <span class="math">x^2 + x - 2 = 0</span></p>
<p>(c) <span class="math">x^2 - x - 2 = 0</span></p>
<p>(d) <span class="math">x^2 + x + 2 = 0</span></p>
Step-by-Step Solution
Key Concept: Since α is a 7th root of unity (α^7 = 1), we use the property that 1 + α + α² + ... + α⁶ = 0. We partition the non-trivial roots into two groups and use Vieta's formulas to construct the quadratic equation.
<p><strong>Step 1:</strong> Recognize that α = cos(2π/7) + i·sin(2π/7) = e^(2πi/7) is a primitive 7th root of unity, so α⁷ = 1.</p><p><strong>Step 2:</strong> From the property of 7th roots of unity: 1 + α + α² + α³ + α⁴ + α⁵ + α⁶ = 0.</p><p><strong>Step 3:</strong> Identify β and γ:
• β = α + α² + α⁴
• γ = α³ + α⁵ + α⁶</p><p><strong>Step 4:</strong> Calculate β + γ = α + α² + α³ + α⁴ + α⁵ + α⁶ = -1 (from Step 2, since the sum of all seven roots equals 0)</p><p><strong>Step 5:</strong> Calculate βγ:
βγ = (α + α² + α⁴)(α³ + α⁵ + α⁶)
= α⁴ + α⁶ + α⁷ + α⁵ + α⁷ + α⁸ + α⁷ + α⁹ + α¹⁰
Using α⁷ = 1, reduce all powers modulo 7:
= α⁴ + α⁶ + 1 + α⁵ + 1 + α + 1 + α² + α³
= 3 + (α + α² + α³ + α⁴ + α⁵ + α⁶)
= 3 + (-1) = 2</p><p><strong>Step 6:</strong> Apply Vieta's formulas. For a quadratic with roots β and γ:
x² - (β + γ)x + βγ = 0
x² - (-1)x + 2 = 0
x² + x + 2 = 0</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D