Complex Numbers
Locus of Complex Numbers
Grade 11
Question:
<p>All the points in the set <span>\(S = \left\{\dfrac{\alpha+i}{\alpha-i}; \alpha \in R\right\}\)</span> <span>\((i=\sqrt{-1})\)</span> lie on a</p>
<p>Straight line whose slope is 1.</p>
<p>Circle whose radius is 1.</p>
<p>Circle whose radius is <span>\(\sqrt{2}\)</span>.</p>
<p>Straight line whose slope is <span>\(-1\)</span>.</p>
Step-by-Step Solution
Key Concept: Rationalize the complex number by multiplying by the conjugate of the denominator, then analyze the real and imaginary parts to identify the geometric locus as a circle or line in the complex plane.
<p><strong>Step 1:</strong> Let z = (α + i)/(α - i) where α ∈ ℝ. Rationalize by multiplying numerator and denominator by the conjugate (α + i):</p><p>z = (α + i)²/[(α - i)(α + i)] = (α² + 2αi - 1)/(α² + 1)</p><p><strong>Step 2:</strong> Separate into real and imaginary parts:</p><p>z = (α² - 1)/(α² + 1) + i·(2α)/(α² + 1)</p><p>Let x = (α² - 1)/(α² + 1) and y = 2α/(α² + 1)</p><p><strong>Step 3:</strong> Find the constraint by computing x² + y²:</p><p>x² + y² = [(α² - 1)² + (2α)²]/(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 4:</strong> Therefore x² + y² = 1, which is the equation of a circle centered at the origin with radius 1.</p><p>∴ All points lie on the <strong>unit circle</strong> (or circle with center at origin and radius 1)</p>
Correct Answer: B