Complex Numbers
Modulus of complex numbers
Grade 11
Question:
<p><strong>For Problems 1–4</strong><br>Consider the complex numbers \(z = (1 - i\sin\theta)/(1 + i\cos\theta)\).</p><p><strong>Problem 3.</strong> The value of \(\theta\) for which \(z\) is unimodular is given by</p>
<p>(1) \(n\pi \pm \dfrac{\pi}{6},\, n \in I\)</p>
<p>(2) \(n\pi \pm \dfrac{\pi}{3},\, n \in I\)</p>
<p>(3) \(n\pi \pm \dfrac{\pi}{4},\, n \in I\)</p>
<p>(4) no real values of \(\theta\)</p>
Step-by-Step Solution
Key Concept: A complex number is unimodular when |z| = 1. For z = (1 - i·sin θ)/(1 + i·cos θ), we need |numerator| = |denominator|, which gives us the constraint that determines θ.
<p><strong>Step 1:</strong> For z to be unimodular, we need |z| = 1.</p><p><strong>Step 2:</strong> Calculate |z| = |1 - i·sin θ|/|1 + i·cos θ|</p><p><strong>Step 3:</strong> |1 - i·sin θ| = √(1² + sin²θ) = √(1 + sin²θ)</p><p><strong>Step 4:</strong> |1 + i·cos θ| = √(1² + cos²θ) = √(1 + cos²θ)</p><p><strong>Step 5:</strong> For |z| = 1: √(1 + sin²θ) = √(1 + cos²θ)</p><p><strong>Step 6:</strong> Squaring both sides: 1 + sin²θ = 1 + cos²θ</p><p><strong>Step 7:</strong> This gives sin²θ = cos²θ, so |sin θ| = |cos θ|</p><p><strong>Step 8:</strong> Therefore θ = nπ/4 + π/4 where n is any integer, or equivalently θ = (2n+1)π/4</p><p>∴ Answer: D</p>
Correct Answer: D