<p>If \(z_1\) and \(z_2\) are the complex roots of the equation \((x-3)^3 + 1 = 0\), then \(z_1 + z_2\) equals</p>
Step-by-Step Solution
Key Concept: Expand (x-3)³ + 1 = 0 to get a cubic equation, then use Vieta's formulas to find the sum of roots. The key is recognizing that the sum of all three roots equals the negative of the coefficient of x² divided by the leading coefficient, and subtracting the real root.
<p><strong>Step 1:</strong> Solve (x-3)³ + 1 = 0, which gives (x-3)³ = -1</p><p>Taking cube roots: x - 3 = ∛(-1) × (cube roots of unity)</p><p>The three cube roots of -1 are: -1, (-1)ω, (-1)ω² where ω = e^(2πi/3)</p><p><strong>Step 2:</strong> Therefore the three roots are: x₁ = 2, x₂ = 3 - ω, x₃ = 3 - ω²</p><p>where x₁ = 2 is the real root (since -1 is real)</p><p><strong>Step 3:</strong> The complex roots are z₁ = 3 - ω and z₂ = 3 - ω²</p><p>z₁ + z₂ = (3 - ω) + (3 - ω²) = 6 - (ω + ω²)</p><p><strong>Step 4:</strong> Using the property 1 + ω + ω² = 0, we get ω + ω² = -1</p><p>∴ z₁ + z₂ = 6 - (-1) = <strong>7</strong></p>
Correct Answer: D