Complex Numbers
Locus of Complex Numbers
Grade 11
Question:
<p>If <i>|z - 2 - i| = |z| sin</i>\(\left(\frac{\pi}{4} - \arg z\right)\), where <i>i = √−1</i>, then locus of <i>z</i>, is</p>
<p>(a) a pair of straight lines</p>
<p>(b) circle</p>
<p>(c) parabola</p>
<p>(d) ellipse</p>
Step-by-Step Solution
Key Concept: Convert the given condition involving |z - 2 - i| and arg z into a standard locus equation. The right side uses the sine of an angle related to arg z, which can be expanded using trigonometric identities and the polar form of complex numbers.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ. Then |z - 2 - i| = √[(x-2)² + (y-1)²]</p><p><strong>Step 2:</strong> Let arg(z) = θ. Then |z| = √(x² + y²) and sin(π/4 - θ) = sin(π/4)cos(θ) - cos(π/4)sin(θ) = (1/√2)[cos(θ) - sin(θ)]</p><p><strong>Step 3:</strong> Since z = |z|e^(iθ), we have cos(θ) = x/|z| and sin(θ) = y/|z|. Therefore: sin(π/4 - θ) = (1/√2)[x/|z| - y/|z|] = (1/√2) · (x-y)/|z|</p><p><strong>Step 4:</strong> Substitute into the given equation: √[(x-2)² + (y-1)²] = |z| · (1/√2) · (x-y)/|z| = (x-y)/√2</p><p><strong>Step 5:</strong> Therefore: √[(x-2)² + (y-1)²] = (x-y)/√2</p><p><strong>Step 6:</strong> Square both sides: (x-2)² + (y-1)² = (x-y)²/2</p><p><strong>Step 7:</strong> Expand: x² - 4x + 4 + y² - 2y + 1 = (x² - 2xy + y²)/2</p><p><strong>Step 8:</strong> Multiply by 2: 2x² - 8x + 8 + 2y² - 4y + 2 = x² - 2xy + y²</p><p><strong>Step 9:</strong> Rearrange: x² + y² + 2xy - 8x - 4y + 10 = 0, which gives (x + y)² - 8x - 4y + 10 = 0</p><p><strong>Step 10:</strong> Rewrite as: (x + y - 4)² = 6x + 2y - 6 = 2(3x + y - 3). After completing the square properly, this represents a parabola of the form (x + y - a)² = 2p(3x + y - b) for appropriate constants.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C