Complex Numbers
Locus in complex plane
Grade 11
Question:
<p><b>For Problems 26–28:</b> Complex numbers \(z\) satisfy the equation \(|z - (4/z)| = 2\).</p><p>Locus of \(z\) if \(|z - z_1| = |z - z_2|\), where \(z_1\) and \(z_2\) are complex numbers with the greatest and the least moduli, is</p>
<p>(1) line parallel to the real axis</p>
<p>(2) line parallel to the imaginary axis</p>
<p>(3) line having a positive slope</p>
<p>(4) line having a negative slope</p>
Step-by-Step Solution
Key Concept: Find the extreme moduli of z from |z - 4/z| = 2 by recognizing this represents points equidistant from z and 4/z, then the locus |z - z₁| = |z - z₂| is the perpendicular bisector of the line segment joining the points with maximum and minimum |z|.
<p><strong>Step 1:</strong> Analyze |z - 4/z| = 2. Let z = re^(iθ). Then |z - 4/z| = |z - 4·z̄/|z|²| represents the distance condition.</p><p><strong>Step 2:</strong> For real z > 0, we have |z - 4/z| = 2. This gives us z - 4/z = ±2. Solving z² - 2z - 4 = 0 yields z = 1 ± √5. Thus z₁ = 1 + √5 (maximum modulus) and z₂ = 1 - √5 (but we take the reciprocal relationship: moduli are 1 + √5 and -1 + √5 = √5 - 1).</p><p><strong>Step 3:</strong> The maximum modulus is |z₁| = 1 + √5 and minimum modulus is |z₂| = √5 - 1. Note that |z₁| · |z₂| = (1 + √5)(√5 - 1) = 4, confirming the constraint geometry.</p><p><strong>Step 4:</strong> The locus of points equidistant from z₁ and z₂ (where z₁ = 1 + √5 and z₂ = √5 - 1 on the real axis) is the perpendicular bisector, which is a vertical line at x = (z₁ + z₂)/2 = (1 + √5 + √5 - 1)/2 = √5.</p><p>∴ Answer: A (The locus is the line Re(z) = √5, or equivalently x = √5)</p>
Correct Answer: A