Complex Numbers
Locus in Complex Plane
Grade 11
Question:
<p>All the points in the set \(S = \left\{\dfrac{\alpha + i}{\alpha - i}; \alpha \in \mathbb{R}\right\}\) \((i = \sqrt{-1})\) lie on a <strong>(JEE Main 2019, April)</strong></p>
<p>Straight line whose slope is 1.</p>
<p>Circle whose radius is 1.</p>
<p>Circle whose radius is \(\sqrt{2}\).</p>
<p>Straight line whose slope is \(-1\).</p>
Step-by-Step Solution
Key Concept: Convert the complex number z = (α+i)/(α-i) to standard form by multiplying by the conjugate, then eliminate the parameter α to find the locus equation. The result will be a circle (or part of it) in the complex plane.
<p><strong>Step 1:</strong> Let z = (α+i)/(α-i) where α ∈ ℝ. Multiply numerator and denominator by the conjugate (α+i):</p><p>z = (α+i)(α+i)/[(α-i)(α+i)] = (α+i)²/(α²+1)</p><p><strong>Step 2:</strong> Expand (α+i)² = α² - 1 + 2αi</p><p>z = (α²-1)/(α²+1) + i·(2α)/(α²+1)</p><p><strong>Step 3:</strong> Let z = x + iy. Then:</p><p>x = (α²-1)/(α²+1) and y = 2α/(α²+1)</p><p><strong>Step 4:</strong> Compute x² + y²:</p><p>x² + y² = [(α²-1)/(α²+1)]² + [2α/(α²+1)]² = [(α²-1)² + 4α²]/(α²+1)²</p><p>= [α⁴ - 2α² + 1 + 4α²]/(α²+1)² = [α⁴ + 2α² + 1]/(α²+1)² = (α²+1)²/(α²+1)² = 1</p><p><strong>Step 5:</strong> Therefore x² + y² = 1, which is a circle with center (0,0) and radius 1.</p><p><strong>Note:</strong> The point (1, 0) is excluded since it would require α = i ∉ ℝ.</p><p>∴ All points lie on the <strong>unit circle |z| = 1</strong> (excluding the point z = 1)</p>
Correct Answer: B