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 real 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) \((c\bar{a} - a\bar{c})^2 = (b\bar{c} - c\bar{b})(\bar{a}b - \bar{b}a)\)</p>
Step-by-Step Solution
Key Concept: If a quadratic with complex coefficients has one purely real root r, then substituting r into the equation gives a+b/r+c/r²=0 (after dividing by r²≠0), which means the real and imaginary parts must separately satisfy specific algebraic relationships between a, b, c.
<p><strong>Step 1:</strong> Let r be the purely real root. Then ar² + br + c = 0.</p><p><strong>Step 2:</strong> Write a = a₁ + ia₂, b = b₁ + ib₂, c = c₁ + ic₂ where a₁,a₂,b₁,b₂,c₁,c₂ ∈ ℝ.</p><p><strong>Step 3:</strong> Substitute: (a₁ + ia₂)r² + (b₁ + ib₂)r + (c₁ + ic₂) = 0</p><p><strong>Step 4:</strong> Separate real and imaginary parts:</p><p>Real: a₁r² + b₁r + c₁ = 0</p><p>Imaginary: a₂r² + b₂r + c₂ = 0</p><p><strong>Step 5:</strong> The condition for one purely real root is that these two equations are satisfied simultaneously. This occurs when the determinant condition: <strong>b² - 4ac = 0</strong> (the standard discriminant condition extends to complex coefficients when seeking real roots).</p><p>Alternatively, the condition is: <strong>arg(b) = arg(2a) and arg(c) = arg(a)</strong>, or equivalently, <strong>b/(2a) and c/a are positive real numbers</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C