<p><strong>For Problems 9–11</strong><br>Let \(z\) be a complex number satisfying \(z^2 + 2z\lambda + 1 = 0\), where \(\lambda\) is a parameter which can take any real value.</p><p><strong>Problem 9.</strong> The roots of this equation lie on a certain circle if</p>
<p>(1) \(-1 < \lambda < 1\)</p>
<p>(2) \(\lambda > 1\)</p>
<p>(3) \(\lambda < 1\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: For a quadratic z² + 2λz + 1 = 0 with real parameter λ, the locus of roots forms a circle because the product of roots is always 1 (constant), which is the defining property of a circle in the complex plane centered at origin.
<p><strong>Step 1:</strong> Apply Vieta's formulas to z² + 2λz + 1 = 0</p><p>Product of roots: z₁ · z₂ = 1 (constant, independent of λ)</p><p>Sum of roots: z₁ + z₂ = -2λ (varies with λ)</p><p><strong>Step 2:</strong> For a root z = x + iy, use the constraint z₁ · z₂ = 1</p><p>If z is a root for some λ, then: z · z₂ = 1, so z₂ = 1/z</p><p><strong>Step 3:</strong> From sum of roots: z + 1/z = -2λ</p><p>Substituting z = x + iy: (x + iy) + (x - iy)/(x² + y²) = -2λ</p><p>For this to hold for all real λ, the imaginary part must equal zero and real part varies.</p><p><strong>Step 4:</strong> The locus satisfies |z|² - 2Re(z)·cos(α) = 1, which represents a circle</p><p>More directly: from z · z̄₂ = 1 and varying λ, all roots lie on |z| = 1 (unit circle)</p><p>∴ Answer: A (The roots lie on the unit circle |z| = 1)</p>
Correct Answer: A