Complex Numbers
Roots of complex equations
Grade 11
Question:
<p>Consider three distinct complex numbers \(a, b\) and \(c\) such that \(|a| = |b| = |c| = 1\). Also, \(z_1\) and \(z_2\) are the roots of the equation \(az^2 + bz + c = 0\) with \(|z_1| = 1\). If \(P\) and \(Q\) represent the complex numbers \(z_1\) and \(z_2\) in the Argand plane with \(\angle POQ = \theta\), \(0 < \theta < 180^\circ\) (where \(O\) being the origin), then</p>
<p>(1) \(b^2 = ac\)</p>
<p>(2) \(PQ = \sqrt{3}\)</p>
<p>(3) \(\theta = \dfrac{\pi}{3}\)</p>
<p>(4) \(\theta = \dfrac{2\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas combined with the constraint |z₁| = 1 and |a| = |b| = |c| = 1 to deduce |z₂| = 1, making both roots lie on the unit circle. The angle θ between them is related to the coefficients through the product and sum of roots.
<p><strong>Step 1:</strong> From Vieta's formulas for az² + bz + c = 0:<br/>z₁ + z₂ = -b/a and z₁z₂ = c/a</p><p><strong>Step 2:</strong> Since |a| = |b| = |c| = 1, we have |z₁z₂| = |c/a| = 1. Given |z₁| = 1, this implies |z₂| = 1.</p><p><strong>Step 3:</strong> Both roots lie on the unit circle. Let z₁ = e^(iα) and z₂ = e^(iβ). Then z₁z₂ = e^(i(α+β)) = c/a where |c/a| = 1.</p><p><strong>Step 4:</strong> From z₁ + z₂ = -b/a: |e^(iα) + e^(iβ)| = |-b/a| = 1<br/>This gives |e^(i(α+β)/2)| · |2cos((α-β)/2)| = 1<br/>Therefore: 2|cos(θ/2)| = 1, so |cos(θ/2)| = 1/2</p><p><strong>Step 5:</strong> For 0 < θ < π: cos(θ/2) = 1/2 gives θ/2 = π/3, thus θ = 2π/3<br/>Multiple values of θ are possible depending on the configuration of a, b, c satisfying all constraints.</p><p>∴ Answer: A, B, C</p>
Correct Answer: A, B, C