<p>If \(\omega\) is the imaginary cube root of unity, then the number of pairs of integers \((a, b)\) such that \(|a\omega + b| = 1\) is ___.</p>
Step-by-Step Solution
Key Concept: Use the property that ω satisfies ω³ = 1 and 1 + ω + ω² = 0, then express |aω + b|² = 1 as a quadratic Diophantine equation in integers a and b.
<p><strong>Step 1:</strong> Let ω = e^(2πi/3) = -1/2 + i√3/2 (primitive cube root of unity).</p><p><strong>Step 2:</strong> Express |aω + b|²:<br/>|aω + b|² = (a·(-1/2) + b)² + (a·√3/2)²<br/>= (-a/2 + b)² + 3a²/4<br/>= a²/4 - ab + b² + 3a²/4<br/>= a² - ab + b²</p><p><strong>Step 3:</strong> Set up the equation:<br/>For |aω + b| = 1, we need: a² - ab + b² = 1</p><p><strong>Step 4:</strong> Solve the Diophantine equation a² - ab + b² = 1:<br/>Rewrite as: (a - b/2)² + 3b²/4 = 1<br/>This gives: (2a - b)² + 3b² = 4</p><p><strong>Step 5:</strong> Find integer solutions by testing values:<br/>• b = 0: (2a)² = 4 → a = ±1 → pairs (1,0), (-1,0)<br/>• b = ±1: (2a - b)² + 3 = 4 → (2a - b)² = 1 → 2a - b = ±1<br/> - If b = 1: a ∈ {0, 1} → pairs (0,1), (1,1)<br/> - If b = -1: a ∈ {0, -1} → pairs (0,-1), (-1,-1)<br/>• |b| ≥ 2: 3b² ≥ 12 > 4, no solutions</p><p><strong>Step 6:</strong> Verify all solutions satisfy the original equation:<br/>(1,0), (-1,0), (0,1), (1,1), (0,-1), (-1,-1) all satisfy a² - ab + b² = 1</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: 6