Complex Numbers
Roots of Complex Equations
Grade None
Question:
<p>Let \(\alpha, \beta\) be real and \(z\) be a complex number. If \(z^2 + \alpha z + \beta = 0\) has two distinct roots on the line \(\text{Re}(z) = 1\), then it is necessary that</p>
<p>\(\beta \in (1, \infty)\)</p>
<p>\(\beta \in (0, 1)\)</p>
<p>\(\beta \in (-1, 0)\)</p>
<p>\(|\beta| = 1\)</p>
Step-by-Step Solution
Key Concept: If a quadratic with real coefficients has two distinct roots on the vertical line Re(z) = 1, then both roots must be complex conjugates of the form 1±ki, which forces the sum of roots to equal 2 and the product to be positive.
<p><strong>Step 1:</strong> Since α, β are real, complex roots must be conjugate pairs. If both roots lie on Re(z) = 1, let them be z₁ = 1 + ki and z₂ = 1 - ki (where k ∈ ℝ, k ≠ 0 for distinct roots).</p><p><strong>Step 2:</strong> By Vieta's formulas:</p><p>Sum: z₁ + z₂ = (1 + ki) + (1 - ki) = 2 = -α</p><p>Therefore: <strong>α = -2</strong></p><p><strong>Step 3:</strong> Product: z₁·z₂ = (1 + ki)(1 - ki) = 1 + k² = β</p><p>Since k² > 0 for distinct roots: <strong>β > 1</strong></p><p><strong>Step 4:</strong> Also verify discriminant: Δ = α² - 4β = 4 - 4β < 0, confirming complex roots (since β > 1).</p><p>∴ Answer: The necessary condition is <strong>α = -2 and β > 1</strong> (or equivalent such as β > 1, α² = 4 < 4β)</p>
Correct Answer: A