Complex Numbers
Roots of unity and modulus
Grade 11
Question:
<p>For any integer <em>k</em>, let \(\alpha_k = \cos\dfrac{k\pi}{7} + i\sin\dfrac{k\pi}{7}\), where \(i = \sqrt{-1}\). Value of the expression \(\dfrac{\displaystyle\sum_{k=1}^{12}|\alpha_{k+1} - \alpha_k|}{\displaystyle\sum_{k=1}^{3}|\alpha_{4k-1} - \alpha_{4k-2}|}\) is _______.</p><p>(JEE Advanced 2015)</p>
Step-by-Step Solution
Key Concept: Recognize that αₖ = e^(ikπ/7) lies on the unit circle, so |αₖ₊₁ - αₖ| = 2sin(π/14) for consecutive points. Use periodicity: α₁₄ = α₀ = 1, and the numerator sum completes ~1.7 cycles around the circle.
<p><strong>Step 1:</strong> Express αₖ in exponential form: αₖ = e^(ikπ/7). These points lie on the unit circle at angles kπ/7.</p><p><strong>Step 2:</strong> For consecutive points, |αₖ₊₁ - αₖ| = |e^(i(k+1)π/7) - e^(ikπ/7)| = |e^(ikπ/7)||e^(iπ/7) - 1| = |e^(iπ/7) - 1| = 2sin(π/14) for all k where both lie on the principal arc.</p><p><strong>Step 3:</strong> Evaluate numerator: Since α₁₄ = e^(i·14π/7) = e^(2πi) = 1 = α₀, the sequence has period 14. For k=1 to 12: all 12 terms equal 2sin(π/14). Sum = 12·2sin(π/14).</p><p><strong>Step 4:</strong> Evaluate denominator: Calculate the three terms:</p><p>• k=1: |α₃ - α₂| = 2sin(π/14)</p><p>• k=2: |α₇ - α₆| = 2sin(π/14)</p><p>• k=3: |α₁₁ - α₁₀| = 2sin(π/14)</p><p>Sum = 3·2sin(π/14).</p><p><strong>Step 5:</strong> Calculate ratio: (12·2sin(π/14))/(3·2sin(π/14)) = 12/3 = 4.</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4