Question

If \( z^2 + z + 1 = 0 \), where \( z \) is a complex number, then the value of \[ \left( z + \frac{1}{z} \right)^2 + \left( z^2 + \frac{1}{z^2} \right)^2 + \left( z^3 + \frac{1}{z^3} \right)^2 + \cdots + \left( z^6 + \frac{1}{z^6} \right)^2 \] is

• A. \( 18 \)
• B. \( 54 \)
• C. \( 6 \)
• D. \( 19 \)
• E. \( 12 \)

Hint:

If \( z^2 + z + 1 = 0 \), then \( z \) is a cube root of unity and powers repeat every 3 terms.

Answer 1

The Correct Option is C

Concept: If \( z \) satisfies \( z^2 + z + 1 = 0 \), then: \[ z^3 = 1 \text{and} z \neq 1 \] So \( z \) is a complex cube root of unity. Also, \[ \frac{1}{z} = z^2 \]

Step 1: Find \( \frac{1}{z} \). From: \[ z^2 + z + 1 = 0 \Rightarrow z^2 = -z - 1 \] Divide by \(z\): \[ z + 1 + \frac{1}{z} = 0 \Rightarrow \frac{1}{z} = -z - 1 \]

Step 2: Use identity \( z^3 = 1 \). Thus powers repeat: \[ z^1 = z, z^2 = z^2, z^3 = 1, z^4 = z, z^5 = z^2, z^6 = 1 \] Similarly: \[ \frac{1}{z^n} = z^{3-n} \]

Step 3: Compute each term. \[ z + \frac{1}{z} = z + z^2 = -1 \Rightarrow (-1)^2 = 1 \] \[ z^2 + \frac{1}{z^2} = z^2 + z = -1 \Rightarrow 1 \] \[ z^3 + \frac{1}{z^3} = 1 + 1 = 2 \Rightarrow 4 \] \[ z^4 + \frac{1}{z^4} = z + z^2 = -1 \Rightarrow 1 \] \[ z^5 + \frac{1}{z^5} = z^2 + z = -1 \Rightarrow 1 \] \[ z^6 + \frac{1}{z^6} = 1 + 1 = 2 \Rightarrow 4 \]

Step 4: Add all values. \[ 1 + 1 + 4 + 1 + 1 + 4 = 12 \]