Complex Numbers
Quadratic equations with complex coefficients
Grade 11
Question:
<p><b>For Problems 14–16:</b> Consider a quadratic equation \(az^2 + bz + c = 0\), where \(a, b, c\) are complex numbers.</p><p>The condition that the equation has one purely imaginary root is</p>
<p>(1) \((c\bar{a} - a\bar{c})^2 = -(b\bar{c} + c\bar{b})(\bar{a}b + \bar{b}a)\)</p>
<p>(2) \((c\bar{a} + a\bar{c})^2 = (b\bar{c} + c\bar{b})(\bar{a}b + \bar{b}a)\)</p>
<p>(3) \((c\bar{a} - a\bar{c})^2 = (b\bar{c} - c\bar{b})(\bar{a}b - \bar{b}a)\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: If z = ki (purely imaginary) is a root, substitute it into az² + bz + c = 0 to get a(-k²) + b(ki) + c = 0, then separate real and imaginary parts to establish a relationship between a, b, c that must hold independently of k's value.
<p><strong>Step 1:</strong> Let z = ki where k ∈ ℝ and k ≠ 0 be a purely imaginary root.</p><p><strong>Step 2:</strong> Substitute into az² + bz + c = 0:<br/>a(ki)² + b(ki) + c = 0<br/>-ak² + bki + c = 0</p><p><strong>Step 3:</strong> Separate into real and imaginary parts:<br/>Real part: -ak² + c = 0<br/>Imaginary part: bk = 0</p><p><strong>Step 4:</strong> For a purely imaginary root to exist independently of k's value, we need the imaginary part coefficient to vanish: <strong>b = 0</strong></p><p><strong>Step 5:</strong> With b = 0, the equation becomes az² + c = 0, which has purely imaginary roots when a and c have opposite signs (or arg(c/a) = π).</p><p><strong>Step 6:</strong> The condition is: <strong>b² = 4ac</strong> becomes degenerate, and the necessary condition is <strong>b = 0</strong> with c/a being a negative real number, OR more generally: <strong>a·c̄ + ā·c is real and opposite in sign to 2|a||c|</strong></p><p>∴ Answer: The condition that one root is purely imaginary is <strong>b = 0 and arg(c/a) = π (equivalently, c/a < 0 as reals)</strong></p>
Correct Answer: A