<p>Find the principal argument of the complex number \(\sin\dfrac{6\pi}{5} + i\left(1+\cos\dfrac{6\pi}{5}\right)\).</p>
Step-by-Step Solution
Key Concept: Convert the complex number using half-angle formulas: recognize that sin(6π/5) + i(1 + cos(6π/5)) can be rewritten as 2sin(3π/5)[sin(3π/5) + i·cos(3π/5)], then convert to standard form using Euler's relation.
<p><strong>Step 1: Apply half-angle identities</strong></p><p>Let z = sin(6π/5) + i(1 + cos(6π/5))</p><p>Use: sin(θ) = 2sin(θ/2)cos(θ/2) and 1 + cos(θ) = 2cos²(θ/2)</p><p>Here, with θ = 6π/5:</p><p>• sin(6π/5) = 2sin(3π/5)cos(3π/5)</p><p>• 1 + cos(6π/5) = 2cos²(3π/5)</p><p><strong>Step 2: Factor the expression</strong></p><p>z = 2sin(3π/5)cos(3π/5) + i·2cos²(3π/5)</p><p>z = 2cos(3π/5)[sin(3π/5) + i·cos(3π/5)]</p><p><strong>Step 3: Rewrite in standard form</strong></p><p>Note that sin(3π/5) + i·cos(3π/5) = i[cos(3π/5) - i·sin(3π/5)] = i·e^(-i·3π/5)</p><p>Since 3π/5 ∈ (π/2, π): sin(3π/5) > 0, cos(3π/5) < 0</p><p>The argument of sin(3π/5) + i·cos(3π/5) is π/2 + 3π/5 = 11π/10</p><p><strong>Step 4: Calculate principal argument</strong></p><p>Since 2cos(3π/5) < 0 (as 3π/5 is in Q2), it contributes phase π</p><p>arg(z) = π + 11π/10 = 21π/10</p><p>Principal argument = 21π/10 - 2π = π/10</p><p><strong>Correction:</strong> Using arg(sin(3π/5) + i·cos(3π/5)) = arg(ie^(-i·3π/5)) directly gives π/2 - 3π/5 = -π/10, and with factor 2cos(3π/5) < 0: arg(z) = π - π/10 = <strong>9π/10</strong></p>
Correct Answer: 9π/10