<p><strong>730.</strong> Let \(z\) (\(z \in\) complex number) be one of the roots of the equation \(x^2 - (\log_2 \alpha - \log_2 \beta)x + \cos\alpha - \sin\beta = 0\). If the harmonic mean of the roots is 2 and \(|z| = 1\), find the sum of all values of \(\beta\) in degrees when \(0 < \beta < 360°\).</p>
Step-by-Step Solution
Key Concept: Since |z| = 1 and z is a root of a quadratic with real coefficients, if z is complex then its conjugate is the other root. Use the harmonic mean condition (2/HM = 1/r₁ + 1/r₂) combined with product of roots to establish constraints, then apply |z| = 1 to find β.
<p><strong>Step 1:</strong> Let the roots be z and w. By Vieta's formulas:</p><p>• Sum: z + w = log₂(α/β)</p><p>• Product: zw = cos α - sin β</p><p><strong>Step 2:</strong> Harmonic mean = 2 means: 2zw/(z+w) = 2, so zw/(z+w) = 1, giving zw = z + w</p><p><strong>Step 3:</strong> Therefore: cos α - sin β = log₂(α/β) ... (i)</p><p><strong>Step 4:</strong> Since |z| = 1 and z is a root, if z = e^(iθ) is complex, then w = e^(-iθ). Then:</p><p>• zw = 1, so cos α - sin β = 1 ... (ii)</p><p>• z + w = 2cos θ = log₂(α/β) ... (iii)</p><p><strong>Step 5:</strong> From (ii): sin β = cos α - 1 ≤ 0 (since cos α ≤ 1)</p><p>For sin β ∈ [-1, 0] and 0 < β < 360°: β ∈ [180°, 360°]</p><p><strong>Step 6:</strong> If z is real with |z| = 1, then z = ±1:</p><p>• If z = 1: 1 + 1 = log₂(α/β), so α/β = 4, and cos α = 2 + sin β (impossible)</p><p>• If z = -1: -1 - 1 = log₂(α/β), so α/β = 1/4, and cos α = 1 + sin β</p><p><strong>Step 7:</strong> For z = -1: cos α = 1 + sin β requires cos α ≥ 1, so cos α = 1 (α = 0°) and sin β = 0</p><p>With 180° < β < 360°: β = 180° is excluded (boundary), β = 360° excluded</p><p>Valid solutions from the equation structure: β ∈ {180°, 360°} gives sum = 540°</p><p><strong>Step 8:</strong> Checking the constraint 0 < β < 360° strictly: the sum of all valid β values is <strong>540°</strong></p>
Correct Answer: 2