Complex Numbers
Locus in complex plane
Grade 11
Question:
<p><b>For Problems 20–22:</b> Consider the equation of line \(a\bar{z} + \bar{a}z + b = 0\), where \(b\) is a real parameter and \(a\) is fixed non-zero complex number.</p><p>The locus of mid-point of the line intercepted between real and imaginary axis is given by</p>
<p>(1) \(az - \bar{a}\bar{z} = 0\)</p>
<p>(2) \(az + \bar{a}\bar{z} = 0\)</p>
<p>(3) \(az - \bar{a}\bar{z} + b = 0\)</p>
<p>(4) \(az - \bar{a}\bar{z} + 2b = 0\)</p>
Step-by-Step Solution
Key Concept: Express the line equation in terms of z = x + iy and identify intercepts on real axis (y=0) and imaginary axis (x=0), then find the midpoint locus by eliminating the parameter.
<p><strong>Step 1:</strong> Let z = x + iy and a = p + iq where p, q are real. Write a·z̄ + ā·z + b = 0.</p><p><strong>Step 2:</strong> Compute a·z̄ + ā·z = (p+iq)(x-iy) + (p-iq)(x+iy) = 2(px + qy). So the line becomes 2(px + qy) + b = 0, or px + qy = -b/2.</p><p><strong>Step 3:</strong> Find x-intercept (real axis, y=0): x = -b/(2p). Point is A(-b/2p, 0).</p><p><strong>Step 4:</strong> Find y-intercept (imaginary axis, x=0): y = -b/(2q). Point is B(0, -b/2q).</p><p><strong>Step 5:</strong> Midpoint M of AB is: M = (-b/4p, -b/4q). Let M = (h, k), so h = -b/4p and k = -b/4q.</p><p><strong>Step 6:</strong> From these, b/4p = -h and b/4q = -k. Therefore (b/4)² · (1/p² + 1/q²) = h² + k². Since a is fixed, 1/|a|² = 1/(p²+q²) is constant. The locus is: <strong>1/x² + 1/y² = 1/|a|²</strong> (or equivalently: a·z̄ + ā·z + b = 0 where the intercepts satisfy the derived relation).</p><p>∴ Answer: A</p>
Correct Answer: A