<p>If \(\alpha\) and \(\beta\) be the roots of the equation \(x^2 - 2x + 2 = 0\), then the least value of \(n\) for which \((\alpha/\beta)^n = 1\) is __________.</p>
Step-by-Step Solution
Key Concept: Express the roots in polar form using Euler's formula, then find when their quotient raised to power n equals 1 by finding the argument of α/β and determining the smallest n that makes the total argument a multiple of 2π.
<p><strong>Step 1:</strong> Find the roots of x² - 2x + 2 = 0 using the quadratic formula:</p><p>x = (2 ± √(4-8))/2 = (2 ± 2i)/2 = 1 ± i</p><p>So α = 1 + i and β = 1 - i</p><p><strong>Step 2:</strong> Convert to polar form:</p><p>|α| = √(1² + 1²) = √2, arg(α) = π/4</p><p>|β| = √2, arg(β) = -π/4</p><p>So α = √2·e^(iπ/4) and β = √2·e^(-iπ/4)</p><p><strong>Step 3:</strong> Calculate α/β:</p><p>α/β = (√2·e^(iπ/4))/(√2·e^(-iπ/4)) = e^(i(π/4 + π/4)) = e^(iπ/2)</p><p><strong>Step 4:</strong> Find least n where (α/β)^n = 1:</p><p>(e^(iπ/2))^n = 1</p><p>e^(inπ/2) = 1</p><p>This requires nπ/2 = 2πk for integer k</p><p>So n = 4k, and the least positive value is n = 4</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4