<p>If \(z^2 + z + 1 = 0\), where \(z\) is a complex number, then the value of \(\left(z + \dfrac{1}{z}\right)^2 + \left(z^2 + \dfrac{1}{z^2}\right)^2 + \left(z^3 + \dfrac{1}{z^3}\right)^2 + \cdots + \left(z^6 + \dfrac{1}{z^6}\right)^2\) is</p>
Step-by-Step Solution
Key Concept: From z² + z + 1 = 0, we get z³ = 1 (z is a primitive cube root of unity), making z⁶ = 1. This periodicity means z^n + 1/z^n repeats with period 3, reducing the sum to just three distinct terms.
<p><strong>Step 1:</strong> From z² + z + 1 = 0, multiply by (z - 1): (z - 1)(z² + z + 1) = 0 ⟹ z³ - 1 = 0 ⟹ z³ = 1. Since z ≠ 1, z is a primitive cube root of unity (ω where ω = e^(2πi/3)).</p><p><strong>Step 2:</strong> Find z + 1/z. Since z³ = 1, we have 1/z = z². From z² + z + 1 = 0: z + z² = -1, so z + 1/z = -1.</p><p><strong>Step 3:</strong> Find z² + 1/z². We have (z + 1/z)² = z² + 2 + 1/z² = 1. Thus z² + 1/z² = -1.</p><p><strong>Step 4:</strong> Find z³ + 1/z³ = 1 + 1 = 2 (since z³ = 1 and 1/z³ = 1).</p><p><strong>Step 5:</strong> Use periodicity: Since z³ = 1, the sequence repeats with period 3:</p><p>• z + 1/z = z⁴ + 1/z⁴ = z⁷ + 1/z⁷ = -1</p><p>• z² + 1/z² = z⁵ + 1/z⁵ = z⁸ + 1/z⁸ = -1</p><p>• z³ + 1/z³ = z⁶ + 1/z⁶ = 2</p><p><strong>Step 6:</strong> Calculate the sum:</p><p>Sum = (-1)² + (-1)² + 2² + (-1)² + (-1)² + 2² = 1 + 1 + 4 + 1 + 1 + 4 = <strong>12</strong></p><p>∴ Answer: D</p>
Correct Answer: D