Complex Numbers
Locus and Lines
Grade 11
Question:
<p>Let <span>a</span> and <span>b</span> be two fixed non-zero complex numbers and <span>z</span> is a variable complex number. If the lines <span>az + \bar{a}\bar{z} + 1 = 0</span> and <span>bz + \bar{b}\bar{z} - 1 = 0</span> are mutually perpendicular, then</p>
<p>(a) <span>ab + \bar{a}\bar{b} = 0</span></p>
<p>(b) <span>ab - \bar{a}\bar{b} = 0</span></p>
<p>(c) <span>\bar{a}b - a\bar{b} = 0</span></p>
<p>(d) <span>\bar{a}b + a\bar{b} = 0</span></p>
Step-by-Step Solution
Key Concept: Lines in the complex plane represented as az + ā·z̄ + c = 0 are perpendicular when the slopes of their corresponding real forms satisfy m₁·m₂ = -1. For complex lines, this translates to a perpendicularity condition involving the coefficients.
<p><strong>Step 1: Convert to standard form</strong></p><p>Let a = a₁ + ia₂ and b = b₁ + ib₂. The equation az + āz̄ + c = 0 can be rewritten. Substituting z = x + iy:</p><p>az + āz̄ = (a₁ + ia₂)(x + iy) + (a₁ - ia₂)(x - iy)</p><p>= a₁x - a₂y + ia₂x + ia₁y + a₁x + a₂y - ia₂x + ia₁y</p><p>= 2a₁x + 2a₁y = 2(a₁x + a₂y)</p><p><strong>Step 2: Identify slopes</strong></p><p>Line 1: az + āz̄ + 1 = 0 gives 2(a₁x + a₂y) + 1 = 0, or a₁x + a₂y = -1/2</p><p>This has slope m₁ = -a₁/a₂</p><p>Line 2: bz + b̄z̄ - 1 = 0 gives 2(b₁x + b₂y) - 1 = 0, or b₁x + b₂y = 1/2</p><p>This has slope m₂ = -b₁/b₂</p><p><strong>Step 3: Apply perpendicularity condition</strong></p><p>For perpendicular lines: m₁·m₂ = -1</p><p>(-a₁/a₂)·(-b₁/b₂) = -1</p><p>a₁b₁/(a₂b₂) = -1</p><p>a₁b₁ = -a₂b₂</p><p><strong>Step 4: Express in terms of complex numbers</strong></p><p>Now, āb + ab̄ = (a₁ - ia₂)(b₁ + ib₂) + (a₁ + ia₂)(b₁ - ib₂)</p><p>= a₁b₁ + ia₁b₂ - ia₂b₁ + a₂b₂ + a₁b₁ - ia₁b₂ + ia₂b₁ + a₂b₂</p><p>= 2a₁b₁ + 2a₂b₂</p><p>For perpendicularity with condition a₁b₁ = -a₂b₂:</p><p>āb + ab̄ = 2a₁b₁ + 2a₂b₂ = 2a₁b₁ + 2(-a₁b₁) = 0</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D