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>If equation has two purely imaginary roots, then which of the following is not true.</p>
<p>(1) \(a\bar{b}\) is purely imaginary</p>
<p>(2) \(b\bar{c}\) is purely imaginary</p>
<p>(3) \(c\bar{a}\) is purely real</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: If a quadratic with complex coefficients has two purely imaginary roots z₁ = iα and z₂ = iβ (where α, β are real), then by Vieta's formulas: sum = -b/a and product = c/a must satisfy specific constraints that reveal which relationship is impossible.
<p><strong>Step 1:</strong> Let the two purely imaginary roots be z₁ = iα and z₂ = iβ where α, β ∈ ℝ.</p><p><strong>Step 2:</strong> By Vieta's formulas:</p><ul><li>Sum: z₁ + z₂ = i(α + β) = -b/a</li><li>Product: z₁z₂ = (iα)(iβ) = -αβ = c/a</li></ul><p><strong>Step 3:</strong> From the sum: b/a must be purely imaginary (of form iγ where γ ∈ ℝ).</p><p><strong>Step 4:</strong> From the product: c/a must be real and non-positive (since -αβ ≤ 0 when α, β have same sign, or positive if opposite signs).</p><p><strong>Step 5:</strong> This eliminates options claiming b/a is real, or c/a is purely imaginary, or a is purely imaginary (which would make c/a complex in incompatible ways).</p><p>∴ Answer: D</p>
Correct Answer: D