Complex Numbers
Locus in complex plane
Grade 11
Question:
<p><b>For Problems 23–25:</b> Consider the equation \(az + b\bar{z} + c = 0\), where \(a, b, c \in \mathbb{Z}\).</p><p>If \(|a| = |b| \neq 0\) and \(\bar{a}c = b\bar{c}\), then \(az + b\bar{z} + c = 0\) represents</p>
<p>(1) an ellipse</p>
<p>(2) a circle</p>
<p>(3) a point</p>
<p>(4) a straight line</p>
Step-by-Step Solution
Key Concept: When |a| = |b| ≠ 0 and ā·c = b·c̄, the equation az + bz̄ + c = 0 must satisfy geometric constraints that force it to represent a line. The condition ā·c = b·c̄ ensures the solution set lies on a real line, not a circle or empty set.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ. Then z̄ = x - iy.</p><p><strong>Step 2:</strong> Substitute into az + bz̄ + c = 0:<br/>a(x + iy) + b(x - iy) + c = 0<br/>(a + b)x + i(a - b)y + c = 0</p><p><strong>Step 3:</strong> Since |a| = |b|, write a = |a|e^(iα) and b = |a|e^(iβ).<br/>The condition ā·c = b·c̄ implies c must have a special form that ensures the imaginary part equals zero for all solutions.</p><p><strong>Step 4:</strong> From ā·c = b·c̄, we get c̄/c = ā/b, meaning arg(c) is constrained such that only real parts remain in the simplified equation.</p><p><strong>Step 5:</strong> This forces the equation to reduce to a real linear equation: (a + b)x + c = 0, which represents a <strong>straight line</strong> (perpendicular to the real axis if a + b ≠ 0, or the entire plane if a + b = 0 with c = 0).</p><p>∴ Answer: D (a straight line)</p>
Correct Answer: D