Complex Numbers
Purely real/imaginary 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 1.</strong> The value of \(\theta\) for which \(z\) is purely real are</p>
<p>(1) \(n\pi - \dfrac{\pi}{4},\, n \in I\)</p>
<p>(2) \(n\pi + \dfrac{\pi}{4},\, n \in I\)</p>
<p>(3) \(n\pi,\, n \in I\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Multiply numerator and denominator by the conjugate of the denominator, then separate real and imaginary parts. A complex number is purely real when its imaginary part equals zero.
<p><strong>Step 1:</strong> Rationalize by multiplying numerator and denominator by the conjugate of denominator (1 - i cos θ):</p><p>z = [(1 - i sin θ)(1 - i cos θ)] / [(1 + i cos θ)(1 - i cos θ)]</p><p><strong>Step 2:</strong> Expand numerator: (1 - i sin θ)(1 - i cos θ) = 1 - i cos θ - i sin θ + i² sin θ cos θ = 1 - i(sin θ + cos θ) - sin θ cos θ = (1 - sin θ cos θ) - i(sin θ + cos θ)</p><p><strong>Step 3:</strong> Expand denominator: (1 + i cos θ)(1 - i cos θ) = 1 - i² cos² θ = 1 + cos² θ</p><p><strong>Step 4:</strong> Therefore: z = [(1 - sin θ cos θ) - i(sin θ + cos θ)] / (1 + cos² θ)</p><p><strong>Step 5:</strong> For z to be purely real, the imaginary part must be zero: sin θ + cos θ = 0, which gives sin θ = -cos θ, or tan θ = -1</p><p><strong>Step 6:</strong> This yields θ = -π/4 + nπ, where n ∈ ℤ (or θ = 3π/4 + nπ)</p><p>∴ Answer: C</p>
Correct Answer: C