Complex Numbers
Roots of Unity
Grade 11
Question:
<p>If <math>1, \omega, \omega^2, \omega^3, \ldots, \omega^{n-1}</math> are <i>n</i> nth roots of unity, then <math>(1 - \omega)(1 - \omega^2)(1 - \omega^3)\cdots(1 - \omega^{n-1})</math> equals</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) <i>n</i></p>
<p>(d) <i>n</i><sup>2</sup></p>
Step-by-Step Solution
Key Concept: Use the polynomial whose roots are the nth roots of unity to establish a relationship. The product (1-ω)(1-ω²)···(1-ω^(n-1)) can be evaluated using the polynomial x^n - 1 = (x-1)(x-ω)(x-ω²)···(x-ω^(n-1)).
<p><strong>Step 1: Recall the polynomial factorization</strong></p><p>Since 1, ω, ω², ..., ω^(n-1) are the nth roots of unity, they satisfy x^n = 1, or equivalently:</p><p>x^n - 1 = (x - 1)(x - ω)(x - ω²)(x - ω³)···(x - ω^(n-1))</p><p><strong>Step 2: Evaluate at x = 1 using L'Hôpital's Rule or direct substitution after factoring</strong></p><p>We can rewrite:</p><p>x^n - 1 = (x - 1)[(x - ω)(x - ω²)···(x - ω^(n-1))]</p><p>Dividing both sides by (x - 1):</p><p>x^(n-1) + x^(n-2) + x^(n-3) + ··· + x + 1 = (x - ω)(x - ω²)(x - ω³)···(x - ω^(n-1))</p><p><strong>Step 3: Substitute x = 1</strong></p><p>LHS: 1 + 1 + 1 + ··· + 1 (n terms) = n</p><p>RHS: (1 - ω)(1 - ω²)(1 - ω³)···(1 - ω^(n-1))</p><p><strong>Step 4: Equate both sides</strong></p><p>(1 - ω)(1 - ω²)(1 - ω³)···(1 - ω^(n-1)) = n</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C