<p>Consider the region \(S\) of complex numbers \(a\) such that \(|z^2 - az + 1| = 1\), where \(|z| = 1\). Then area of \(S\) in the Argand plane is</p>
Step-by-Step Solution
Key Concept: For |z| = 1, write z = e^(iθ) and manipulate |z² - az + 1| = 1 to find the locus of a in the Argand plane. Since z² + 1 = z(a - (z - 1/z)), the condition becomes |a - (z + 1/z)| = 1 for all z on unit circle.
<p><strong>Step 1:</strong> Let |z| = 1, so z = e^(iθ). Then z² - az + 1 = 0 gives us |z² + 1| = |az|, or equivalently |z + 1/z| = |a|.</p><p><strong>Step 2:</strong> Since z = e^(iθ), we have z + 1/z = e^(iθ) + e^(-iθ) = 2cos(θ). As θ varies from 0 to 2π, this traces values in [-2, 2] on the real axis.</p><p><strong>Step 3:</strong> The condition |z² - az + 1| = 1 for all |z| = 1 requires |a - (z + 1/z)| = 1. This means a must be at distance 1 from every point that z + 1/z can reach, i.e., from the interval [-2, 2].</p><p><strong>Step 4:</strong> The set of points at distance exactly 1 from the interval [-2, 2] forms an annulus: the region between circles centered at origin with radii 1 and 3. The outer circle (|a| = 3) is traced when a is distance 1 from the point 2; the inner circle (|a| = 1) is traced when a is distance 1 from the point -2.</p><p><strong>Step 5:</strong> Area of annulus = π(3²) - π(1²) = 9π - π = 8π</p><p>∴ Answer: B</p>
Correct Answer: B