Animated Solution for Mathematics - Complex Numbers: If $z^2+z+1=0,z\in \mathbb{C}$, then $\left|{\sum }_{n=1}^{15}{\left(z^n+(-1{)}^n\frac{1}{z^n}\right)}^2\right|$ is equal to \_\_\_\_\_.

The equation $z^2+z+1=0$ is the fundamental identity for the non-real cube roots of unity, denoted as $\omega$ and ${\omega }^2$. These values represent points on the unit circle at angles of $120^{\circ }$ and $240^{\circ }$.

They satisfy the essential properties:

Consider the expression inside the summation: ${\left(z^n+(-1{)}^n\frac{1}{z^n}\right)}^2$. We apply the algebraic identity $(a+b{)}^2=a^2+b^2+2ab$ to expand this term.

The expansion yields:

Simplifying the terms, we note that $(-1{)}^{2n}=1$ because $2n$ is always even. The middle term simplifies as the $z^n$ factors cancel out.

This reduces our expression to:

The total summation we must evaluate is:

By the linearity of summation, we decompose the problem into three distinct parts. We assume $z=\omega$ for this evaluation.

First, consider ${\sum }_{n=1}^{15}{\omega }^{2n}$. The sequence of terms is ${\omega }^2,{\omega }^4,{\omega }^6,\dots$, which simplifies to ${\omega }^2,\omega ,1,\dots$. This is a repeating cycle of length 3. Since 15 is a multiple of 3, we have 5 full cycles, and the sum is $5\times (1+\omega +{\omega }^2)=0$.

Second, consider ${\sum }_{n=1}^{15}{\omega }^{-2n}$. Since ${\omega }^{-2}=\omega$, this is equivalent to ${\sum }_{n=1}^{15}{\omega }^n$. Again, because 15 is a multiple of 3, the sum of these terms is $0$.

Third, consider ${\sum }_{n=1}^{15}2(-1{)}^n$. This expands to $2\times [(-1)+1-1+1-\cdots -1]$. Because there are 15 terms (an odd number), the pairs cancel out, leaving a single $-1$. Thus, the sum is $2\times (-1)=-2$.

Summing these components together, we find the total value $S$:

The problem asks for the modulus of the result. Therefore: