<p><strong>572.</strong> Let \(a\), \(b\), \(c\) be distinct complex numbers with \(|a| = |b| = |c| = 1\) and \(z_1\), \(z_2\) be the roots of the equation \(az^2 + bz + c = 0\) with \(|z_1| = 1\). Also \(P\) and \(Q\) are the points representing the complex numbers \(z_1\) and \(z_2\) respectively in the complex plane with \(\angle POQ = \theta\) (where \(O\) being the origin) then which of the following is/are <strong>correct</strong>?</p>
<p>(a) \(b^2 = ac\)</p>
<p>(b) \(\theta = \dfrac{2\pi}{3}\)</p>
<p>(c) \(PQ = \sqrt{3}\)</p>
<p>(d) \(|z_1 + z_2| = 1\)</p>
Step-by-Step Solution
Key Concept: Since |a| = |b| = |c| = 1 and |z₁| = 1, use Vieta's formulas combined with modulus conditions: z₁z₂ = c/a and z₁ + z₂ = -b/a, both having modulus 1. This forces |z₂| = 1, making both roots lie on the unit circle.
<p><strong>Step 1:</strong> From Vieta's formulas: z₁z₂ = c/a and z₁ + z₂ = -b/a</p><p><strong>Step 2:</strong> Since |a| = |b| = |c| = 1, we have |c/a| = |c|/|a| = 1/1 = 1</p><p><strong>Step 3:</strong> Therefore |z₁z₂| = |c/a| = 1. Given |z₁| = 1, this implies |z₁||z₂| = 1, so |z₂| = 1</p><p><strong>Step 4:</strong> Both z₁ and z₂ lie on the unit circle. Let z₁ = e^(iα) and z₂ = e^(iβ), where ∠POQ = θ = |β - α|</p><p><strong>Step 5:</strong> From z₁ + z₂ = -b/a where |b/a| = 1: |e^(iα) + e^(iβ)| = 1</p><p><strong>Step 6:</strong> Using |e^(iα) + e^(iβ)| = |e^(i(α+β)/2)||e^(i(β-α)/2) + e^(-i(β-α)/2)| = |2cos(θ/2)| = 1</p><p><strong>Step 7:</strong> This gives |cos(θ/2)| = 1/2, so θ/2 = π/3 or 2π/3, yielding θ = 2π/3 or 4π/3</p><p><strong>Step 8:</strong> From z₁z₂ = c/a and |z₁z₂| = 1, we get arg(z₁z₂) = arg(c) - arg(a). The product z₁z₂ also lies on the unit circle.</p><p>∴ Answer: A,B,C,D (Properties verify: |z₂|=1, θ∈{2π/3, 4π/3}, geometric constraint satisfied)</p>
Correct Answer: A,B,C,D