Key Concepts and Formulas

• Complex Cube Roots of Unity: The roots of $x^3 = 1$ are $1, \omega, \omega^2$, where $\omega = e^{i(2\pi/3)}$ and $\omega^2 = e^{i(4\pi/3)}$.
• Properties of $\omega$: $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.
• The quadratic $z^2 + z + 1 = 0$ has roots $\omega$ and $\omega^2$.

Step-by-Step Solution

Step 1: Identify the roots of the equation We are given $z^2 + z + 1 = 0$. This is a standard quadratic equation. We recognize that its roots are the non-real cube roots of unity, $\omega$ and $\omega^2$. WHY: Multiplying by $(z-1)$ gives $z^3 - 1 = 0$, so $z^3 = 1$. Since $z^2+z+1=0$, $z \neq 1$, implying $z = \omega$ or $z = \omega^2$.

Step 2: Derive a useful relationship Divide the equation $z^2 + z + 1 = 0$ by $z$ (since $z \neq 0$). $\frac{z^2}{z} + \frac{z}{z} + \frac{1}{z} = 0$ $z + 1 + \frac{1}{z} = 0$ $z + \frac{1}{z} = -1$ WHY: This relationship will simplify calculations of the terms in the summation.

Step 3: Simplify the general term Consider the general term $z^k + \frac{1}{z^k}$. Since $z = \omega$ (or $\omega^2$), we can write this as $\omega^k + \frac{1}{\omega^k}$. WHY: We want to find a pattern in these terms so we can evaluate the sum efficiently.

Step 4: Calculate the first few terms Calculate $z^k + \frac{1}{z^k}$ for $k = 1, 2, 3, 4, 5, 6$.

• For $k=1$: $z + \frac{1}{z} = -1$ (from Step 2).
• For $k=2$: $z^2 + \frac{1}{z^2} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$. WHY: $\frac{1}{\omega^2} = \frac{\omega}{\omega^3} = \omega$.
• For $k=3$: $z^3 + \frac{1}{z^3} = \omega^3 + \frac{1}{\omega^3} = 1 + \frac{1}{1} = 2$.
• For $k=4$: $z^4 + \frac{1}{z^4} = \omega^4 + \frac{1}{\omega^4} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1$. WHY: $\omega^4 = \omega^3 \cdot \omega = \omega$.
• For $k=5$: $z^5 + \frac{1}{z^5} = \omega^5 + \frac{1}{\omega^5} = \omega^2 + \frac{1}{\omega^2} = \omega^2 + \omega = -1$. WHY: $\omega^5 = \omega^3 \cdot \omega^2 = \omega^2$.
• For $k=6$: $z^6 + \frac{1}{z^6} = \omega^6 + \frac{1}{\omega^6} = (\omega^3)^2 + \frac{1}{(\omega^3)^2} = 1 + 1 = 2$.

Step 5: Identify the pattern The values of $z^k + \frac{1}{z^k}$ repeat in a cycle of 3: $-1, -1, 2, -1, -1, 2, \dots$ WHY: Since $\omega^3 = 1$, the powers of $\omega$ cycle with period 3.

Step 6: Square the terms Square each term in the sequence: $(-1)^2, (-1)^2, (2)^2, (-1)^2, (-1)^2, (2)^2 = 1, 1, 4, 1, 1, 4$

Step 7: Sum the squared terms The sum is $S = 1 + 1 + 4 + 1 + 1 + 4 = 12$. WHY: This follows directly from the definition of the sum.

Common Mistakes & Tips

• Forgetting the fundamental properties of complex cube roots of unity.
• Not recognizing the pattern in the terms $z^k + \frac{1}{z^k}$.
• Not simplifying powers of $\omega$ using $\omega^3 = 1$.

Summary

The given equation $z^2 + z + 1 = 0$ implies that $z$ is a complex cube root of unity. Using the properties of $\omega$, we found that the terms $z^k + \frac{1}{z^k}$ repeat in a cycle of 3. Squaring these terms and summing them gives the final answer.

Final Answer The final answer is \boxed{12}, which corresponds to option (D).