Complex Numbers
Lines 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 intercept of line on imaginary axis is given by</p>
<p>(1) \(\dfrac{b}{\bar{a} - a}\)</p>
<p>(2) \(\dfrac{2b}{\bar{a} - a}\)</p>
<p>(3) \(\dfrac{b}{2(\bar{a} - a)}\)</p>
<p>(4) \(\dfrac{b}{a - \bar{a}}\)</p>
Step-by-Step Solution
Key Concept: The imaginary axis intercept occurs when the real part of z is zero (z = iy). Substitute z = iy into the line equation and solve for y to find where the line crosses the imaginary axis.
<p><strong>Step 1:</strong> For the imaginary axis, set z = iy where y ∈ ℝ and i is the imaginary unit.</p><p><strong>Step 2:</strong> Substitute z = iy into the equation a̅z + āz̄ + b = 0:</p><p>a̅(iy) + ā(−iy) + b = 0</p><p>ia̅y − iāy + b = 0</p><p><strong>Step 3:</strong> Let a = p + iq where p, q ∈ ℝ. Then a̅ = p − iq and ā = p + iq.</p><p>i(p − iq)y − i(p + iq)y + b = 0</p><p>i(p − iq − p − iq)y + b = 0</p><p>−2iqy + b = 0</p><p><strong>Step 4:</strong> Solve for y:</p><p>y = b/(2iq) = −ib/(2q)</p><p><strong>Step 5:</strong> The imaginary axis intercept is the point (0, −ib/2q) or the intercept length is |b|/(2|Im(a)|).</p><p>∴ Answer: A</p>
Correct Answer: A