Complex Numbers
Roots of Unity
Grade None
Question:
<p>If <em>a</em> = cos(2π/7) + <em>i</em> sin(2π/7), then find the quadratic equation whose roots are α = <em>a</em> + <em>a</em><sup>2</sup> + <em>a</em><sup>4</sup> and β = <em>a</em><sup>3</sup> + <em>a</em><sup>5</sup> + <em>a</em><sup>6</sup>.</p>
<p>\(x^2 + x + 2 = 0\)</p>
<p>\(x^2 + x - 2 = 0\)</p>
<p>\(x^2 - x + 2 = 0\)</p>
<p>\(x^2 - x - 2 = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize that a = e^(2πi/7) is a 7th root of unity (a^7 = 1), so the sum of all roots satisfies a + a² + ... + a⁶ = -1. Partition these 6 roots into two groups (α and β) using properties of cyclotomic polynomials and Galois conjugates.
<p><strong>Step 1:</strong> Since a = e^(2πi/7), we have a^7 = 1. The equation a^7 - 1 = 0 factors as (a-1)(1 + a + a² + a³ + a⁴ + a⁵ + a⁶) = 0.</p><p><strong>Step 2:</strong> Therefore: a + a² + a³ + a⁴ + a⁵ + a⁶ = -1, which means α + β = a + a² + a⁴ + a³ + a⁵ + a⁶ = -1.</p><p><strong>Step 3:</strong> To find αβ, note that α = a + a² + a⁴ and β = a³ + a⁵ + a⁶. Multiply: αβ = (a + a² + a⁴)(a³ + a⁵ + a⁶).</p><p><strong>Step 4:</strong> Expanding: αβ = a⁴ + a⁶ + a⁷ + a⁵ + a⁷ + a⁸ + a⁷ + a⁹ + a¹⁰. Using a⁷ = 1, reduce: αβ = a⁴ + a⁶ + 1 + a⁵ + 1 + a + 1 + a² + a³ = (1 + a + a² + a³ + a⁴ + a⁵ + a⁶) + 2 = -1 + 2 = 2.</p><p><strong>Step 5:</strong> The quadratic equation with sum of roots = -1 and product = 2 is: x² - (α + β)x + αβ = 0, giving x² + x + 2 = 0.</p><p>∴ Answer: <strong>x² + x + 2 = 0</strong> (Option A)</p>
Correct Answer: A