1. Given equation for $z$

We have $z^2+z+1=0.$ Since $z^3-1=(z-1)(z^2+z+1),$ and $z\neq 1$, it follows that $z$ is a non-real cube root of unity. So, $z^3=1, \qquad z^{-1}=z^2.$ Also, $|z|=1.$

2. Simplify the inner expression

We need to evaluate $S=\sum_{n=1}^{15}\left(z^n+(-1)^n\frac{1}{z^n}\right)^2.$ Using $\dfrac{1}{z^n}=z^{-n}=z^{2n}$ (because $z^3=1$), we get $z^n+(-1)^n\frac{1}{z^n}=z^n+(-1)^n z^{2n}.$ Hence $\left(z^n+(-1)^n z^{2n}\right)^2=z^{2n}+2(-1)^n z^{3n}+z^{4n}.$ But $z^{3n}=(z^3)^n=1$, so this becomes $=z^{2n}+2(-1)^n+z^{4n}.$ Now reduce powers modulo $3$: $z^{4n}=z^{3n+n}=z^n.$ Thus $\left(z^n+(-1)^n\frac{1}{z^n}\right)^2=z^n+z^{2n}+2(-1)^n.$

3. Use the identity for cube roots of unity

For $n\not\equiv 0 \pmod 3$, $z^n+z^{2n}=z+z^2=-1.$ For $n\equiv 0 \pmod 3$, $z^n+z^{2n}=1+1=2.$ So each term is:

• If $3\mid n$: $2+2(-1)^n.$
• If $3\nmid n$: $-1+2(-1)^n.$

4. Evaluate according to parity and divisibility by 3

From $1$ to $15$:

• Multiples of $3$: $3,6,9,12,15$

  • Odd multiples: $3,9,15$ give $2+2(-1)=0.$
  • Even multiples: $6,12$ give $2+2(1)=4.$

• Non-multiples of $3$:

  • Odd $n$: $1,5,7,11,13$ give $-1+2(-1)=-3.$
  • Even $n$: $2,4,8,10,14$ give $-1+2(1)=1.$

So, $S=3\cdot 0+2\cdot 4+5\cdot(-3)+5\cdot 1.$ $S=0+8-15+5=-2.$ Therefore, $|S|=|-2|=2.$

5. Final answer

$\boxed{2}$