<p>Let <span class="math">\omega_n = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}</span>, where <span class="math">i = \sqrt{-1}</span>. Then</p>
Step-by-Step Solution
Key Concept: Recognize that ωₙ is a primitive n-th root of unity, and understand that the sum of all n-th roots of unity equals zero. This is because they are roots of xⁿ - 1 = 0, whose coefficients satisfy Vieta's formulas.
<p><strong>Step 1:</strong> Identify that ω_n = cos(2π/n) + i·sin(2π/n) = e^(i·2π/n) is a primitive n-th root of unity.</p><p><strong>Step 2:</strong> The n-th roots of unity satisfy the equation x^n = 1, or equivalently x^n - 1 = 0.</p><p><strong>Step 3:</strong> Factor: x^n - 1 = (x - 1)(x^(n-1) + x^(n-2) + ... + x + 1) = 0.</p><p><strong>Step 4:</strong> The roots are 1, ω_n, ω_n², ..., ω_n^(n-1), where ω_n is the principal n-th root of unity.</p><p><strong>Step 5:</strong> By Vieta's formulas, the sum of all n roots of x^n - 1 = 0 is the negative of the coefficient of x^(n-1) divided by the leading coefficient. Since the coefficient of x^(n-1) is 0, the sum is 0.</p><p><strong>Step 6:</strong> Therefore: 1 + ω_n + ω_n² + ω_n³ + ... + ω_n^(n-1) = 0.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C