Key Concepts and Formulas

• Cube Roots of Unity: The solutions to $z^3 = 1$ are $1, \omega, \omega^2$, where $\omega = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and $\omega^2 = e^{4\pi i/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$.
• Properties of Cube Roots of Unity: $\omega^3 = 1$, $1 + \omega + \omega^2 = 0$, and $\frac{1}{\omega} = \omega^2$.
• Summation Formulas: $\sum_{k=1}^n a_k + b_k = \sum_{k=1}^n a_k + \sum_{k=1}^n b_k$

Step-by-Step Solution

Step 1: Identify the Roots

Since $z^2 + z + 1 = 0$, multiplying by $(z-1)$ gives $z^3 - 1 = 0$, so $z^3 = 1$. Since $z^2 + z + 1 = 0$, $z \neq 1$. Thus, $z = \omega$ or $z = \omega^2$, where $\omega$ and $\omega^2$ are the non-real cube roots of unity. We can choose $z = \omega$ without loss of generality.

Step 2: Simplify the General Term

We need to simplify $\left(z^n + (-1)^n \frac{1}{z^n}\right)^2$. Substituting $z = \omega$, we get $\left(\omega^n + (-1)^n \frac{1}{\omega^n}\right)^2$. Since $\frac{1}{\omega} = \omega^2$, we have $\frac{1}{\omega^n} = (\omega^2)^n = \omega^{2n}$. Thus, the expression becomes $\left(\omega^n + (-1)^n \omega^{2n}\right)^2$.

Expanding the square, we get $(\omega^n)^2 + 2 (\omega^n) ((-1)^n \omega^{2n}) + ((-1)^n \omega^{2n})^2 = \omega^{2n} + 2 (-1)^n \omega^{3n} + (-1)^{2n} \omega^{4n}$. Since $\omega^3 = 1$, we have $\omega^{3n} = (\omega^3)^n = 1^n = 1$. Also, $(-1)^{2n} = 1$. Therefore, the expression simplifies to $\omega^{2n} + 2 (-1)^n + \omega^{4n}$. Since $\omega^{4n} = \omega^{3n} \cdot \omega^n = \omega^n$, the expression becomes $\omega^{2n} + 2(-1)^n + \omega^n$.

Step 3: Evaluate the Summation

We need to find $\left| \sum_{n=1}^{15} \left(\omega^{2n} + \omega^n + 2(-1)^n\right) \right|$. We can split the summation into three parts: $S = \sum_{n=1}^{15} \omega^{2n} + \sum_{n=1}^{15} \omega^n + \sum_{n=1}^{15} 2(-1)^n$.

Step 4: Evaluate the Sum of $\omega^n$

$\sum_{n=1}^{15} \omega^n = \omega + \omega^2 + \omega^3 + \dots + \omega^{15}$. Since $1 + \omega + \omega^2 = 0$, the sum of any three consecutive powers of $\omega$ is zero. Since 15 is a multiple of 3, the sum is zero: $\sum_{n=1}^{15} \omega^n = 0$.

Step 5: Evaluate the Sum of $\omega^{2n}$

$\sum_{n=1}^{15} \omega^{2n} = \omega^2 + \omega^4 + \omega^6 + \dots + \omega^{30}$. Since $\omega^3 = 1$, the powers cycle as $\omega^2, \omega, 1, \omega^2, \omega, 1, \dots$. The sum of each cycle is $\omega^2 + \omega + 1 = 0$. Since 15 is a multiple of 3, the sum is zero: $\sum_{n=1}^{15} \omega^{2n} = 0$.

Step 6: Evaluate the Sum of $2(-1)^n$

$\sum_{n=1}^{15} 2(-1)^n = 2 \sum_{n=1}^{15} (-1)^n = 2(-1 + 1 - 1 + 1 - \dots - 1)$. Since there are 15 terms, the sum is $2(-1) = -2$.

Step 7: Calculate the Final Result

$S = 0 + 0 + (-2) = -2$. The problem asks for the modulus of this sum, which is $|S| = |-2| = 2$.

Common Mistakes & Tips

• Incorrectly applying cube root properties: Make sure you use $\omega^3=1$ and $1+\omega+\omega^2=0$ correctly.
• Forgetting the modulus: Remember to take the absolute value at the end.
• Messing up the alternating sum: Be careful with the sum of $(-1)^n$; it depends on the number of terms.

Summary

By recognizing the roots of the quadratic equation as cube roots of unity, simplifying the general term using the properties of $\omega$, and evaluating the three summations separately, we found that the modulus of the sum is 2.

Final Answer

The final answer is \boxed{2}.