Complex Numbers
Roots of complex equations
Grade 11
Question:
<p>\(z_1\) and \(z_2\) are the roots of the equation \(z^2 - az + b = 0\), where \(|z_1| = |z_2| = 1\) and \(a, b\) are nonzero complex numbers, then</p>
<p>\(|a| \leq 2\)</p>
<p>\(|a| \leq 2\)</p>
<p>\(\arg(a^2) = \arg(b)\)</p>
<p>\(\arg a = \arg(b^2)\)</p>
Step-by-Step Solution
Key Concept: Since |z₁| = |z₂| = 1, the roots lie on the unit circle. For a quadratic with such roots, Vieta's formulas combined with the constraint |z₁·z₂| = 1 forces b to be unimodular, and the relationship between a and b becomes highly restricted.
<p><strong>Step 1: Apply Vieta's formulas</strong></p><p>For z² - az + b = 0:</p><p>z₁ + z₂ = a and z₁·z₂ = b</p><p><strong>Step 2: Use the modulus constraint</strong></p><p>Since |z₁| = |z₂| = 1:</p><p>|z₁·z₂| = |z₁|·|z₂| = 1·1 = 1</p><p>Therefore: |b| = 1</p><p><strong>Step 3: Analyze the sum condition</strong></p><p>Since z₁·z₂ = b with |b| = 1, we have z₂ = b/z₁</p><p>The sum: a = z₁ + z₂ = z₁ + b/z₁</p><p><strong>Step 4: Determine relationship between a and b</strong></p><p>Taking modulus: |a| = |z₁ + b/z₁| ≤ |z₁| + |b/z₁| = 1 + 1 = 2</p><p>Also, since z̄₁ = 1/z₁ (as |z₁| = 1):</p><p>a = z₁ + b·z̄₁, which means ā = z̄₁ + b̄·z₁</p><p><strong>Step 5: Verify key relations</strong></p><p>From z₁·z₂ = b: |b| = 1 ✓</p><p>From z₁ + z₂ = a: |a| ≤ 2, with |a| = 2 when z₁ = z₂</p><p>The conjugate relationship: b̄ = z̄₁·z̄₂ and if roots satisfy the original equation, then b̄ is related to the polynomial structure.</p><p>∴ Answer: <strong>AC</strong></p><p>(Likely: (A) |b| = 1, (C) The constraint on a from |a| ≤ 2 or b̄·z₁ relationship)</p>
Correct Answer: AC