<p>If \(\sqrt{3} + i = (a + ib)(c + id)\), then find the value of \(\tan^{-1}(b/a) + \tan^{-1}(d/c)\).</p>
Step-by-Step Solution
Key Concept: When a complex number equals a product of two complex numbers, the argument of the product equals the sum of individual arguments. Use arg(z₁z₂) = arg(z₁) + arg(z₂) to convert the problem into inverse tangent sum.
<p><strong>Step 1:</strong> Recognize that tan⁻¹(b/a) = arg(a+ib) when a > 0, and tan⁻¹(d/c) = arg(c+id) when c > 0.</p><p><strong>Step 2:</strong> Using the property arg(z₁z₂) = arg(z₁) + arg(z₂), we have:<br>arg(a+ib) + arg(c+id) = arg[(a+ib)(c+id)] = arg(√3 + i)</p><p><strong>Step 3:</strong> Calculate arg(√3 + i). Since √3 + i is in the form r(cos θ + i sin θ) where r = √(3+1) = 2 and tan θ = 1/√3, we get θ = π/6.</p><p><strong>Step 4:</strong> Therefore: tan⁻¹(b/a) + tan⁻¹(d/c) = arg(√3 + i) + 2πn = π/6 + 2πn (accounting for all branches)</p><p><strong>Step 5:</strong> Since inverse tangent has period π for the argument function in general solutions: tan⁻¹(b/a) + tan⁻¹(d/c) = nπ + π/6, where n ∈ ℤ</p><p>∴ Answer: <strong>nπ + π/6, n ∈ ℤ</strong></p>
Correct Answer: nπ + π/6, n ∈ Z