<p>If <em>ω</em> is a complex cube root of unity, then find the value of \((1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \cdots\) to <em>2n</em> factors.</p>
Step-by-Step Solution
Key Concept: Use the fact that ω³ = 1 and 1 + ω + ω² = 0 to reduce powers of ω modulo 3, then recognize the telescoping pattern in the product structure.
<p><strong>Step 1:</strong> Since ω is a primitive cube root of unity, ω³ = 1 and 1 + ω + ω² = 0.</p><p><strong>Step 2:</strong> Reduce all exponents modulo 3. The sequence of exponents is 1, 2, 4, 8, 16, ... which modulo 3 gives: 1, 2, 1, 2, 1, 2, ... (alternating pattern).</p><p><strong>Step 3:</strong> With 2n factors, we have n factors of (1 + ω) and n factors of (1 + ω²).</p><p><strong>Step 4:</strong> Therefore, the product = (1 + ω)ⁿ(1 + ω²)ⁿ = [(1 + ω)(1 + ω²)]ⁿ.</p><p><strong>Step 5:</strong> Calculate (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³ = 1 + ω + ω² + 1 = (1 + ω + ω²) + 1 = 0 + 1 = 1.</p><p><strong>Step 6:</strong> Thus [(1 + ω)(1 + ω²)]ⁿ = 1ⁿ = 1.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1