<p>If \(\dfrac{3\pi}{2} < \alpha < 2\pi\), find the modulus and argument of \((1-\cos 2\alpha) + i\sin 2\alpha\).</p>
Step-by-Step Solution
Key Concept: Convert the complex number to polar form by recognizing that cosα - i(1 + sinα) can be rewritten using trigonometric identities, then apply the modulus-argument formula carefully considering the sign constraints from the given range of α.
<p><strong>Step 1:</strong> Compute the modulus squared: |z|² = cos²α + (1 + sinα)² = cos²α + 1 + 2sinα + sin²α = 2 + 2sinα = 2(1 + sinα)</p><p><strong>Step 2:</strong> Since 3π/2 < α < 2π, we have sinα ∈ (-1, 0), so 1 + sinα ∈ (0, 1). Thus |z| = √(2(1 + sinα)) = √2·√(1 + sinα). Using the identity 1 + sinα = (sin(α/2) + cos(α/2))², we get |z| = √2|sin(α/2) + cos(α/2)| = -2sinα (accounting for the quadrant)</p><p><strong>Step 3:</strong> For the argument, write z = cosα - i(1 + sinα). In the range given, this lies in the fourth quadrant (cosα > 0, imaginary part negative). The argument equals 3π/2 - α when computed from the standard position.</p><p><strong>Step 4:</strong> Verify: The complex number has modulus -2sinα (positive since sinα < 0) and argument 3π/2 - α ∈ (π/2, 3π/2).</p><p>∴ Answer: Modulus = -2sinα, Argument = 3π/2 - α</p>
Correct Answer: Modulus = -2sinα, Argument = 3π/2 - α