<p>Which of the following is equal to \(\displaystyle\sum_{k=1}^{n}\left(\sin\frac{2\pi k}{n} - i\cos\frac{2\pi k}{n}\right)\), where \(i = \sqrt{-1}\)?</p>
Step-by-Step Solution
Key Concept: Recognize that sin(θ) - i·cos(θ) = -i(cos(θ) + i·sin(θ)) = -i·e^(iθ), converting the sum into a geometric series of nth roots of unity.
<p><strong>Step 1:</strong> Rewrite the general term using Euler's formula:</p><p>sin(2πk/n) - i·cos(2πk/n) = -i[cos(2πk/n) + i·sin(2πk/n)] = -i·e^(i·2πk/n)</p><p><strong>Step 2:</strong> Factor out -i from the sum:</p><p>∑(k=1 to n) [-i·e^(i·2πk/n)] = -i·∑(k=1 to n) e^(i·2πk/n)</p><p><strong>Step 3:</strong> Recognize that e^(i·2πk/n) for k = 1, 2, ..., n are the nth roots of unity (excluding 1 when viewed from ω^k where ω = e^(i·2π/n)).</p><p><strong>Step 4:</strong> The sum of all nth roots of unity equals zero:</p><p>∑(k=0 to n-1) e^(i·2πk/n) = 0</p><p>Therefore: ∑(k=1 to n) e^(i·2πk/n) = -1 (since we exclude k=0 term which equals 1, and include k=n term which equals 1)</p><p><strong>Step 5:</strong> Actually, ∑(k=1 to n) e^(i·2πk/n) = ∑(k=1 to n-1) e^(i·2πk/n) + e^(i·2πn/n) = -1 + 1 = 0</p><p><strong>Step 6:</strong> Therefore: -i · 0 = <strong>0</strong></p><p>∴ Answer: B</p>
Correct Answer: B