JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Complex Numbers

The relation $z^2+z+1=0$ makes $z$ a primitive cube root of unity, so $z^3=1$ and $\frac{1}{z}=z^2$. Expanding one term, ${\left(z^n+(-1{)}^n{z}^{-n}\right)}^2=z^{2n}+z^{-2n}+2(-1{)}^n$, the cross term being $2(-1{)}^n{z}^n{z}^{-n}=2(-1{)}^n$. Summing from $n=1$ to $15$, the powers $z^{2n}$ run through repeated blocks $z^2,z^4\equiv z,z^6\equiv 1$ whose per-block sum is $z^2+z+1=0$, and five such blocks give ${\sum }_{n=1}^{15}{z}^{2n}=0$; by the same reasoning ${\sum }_{n=1}^{15}{z}^{-2n}=0$. Only $2{\sum }_{n=1}^{15}(-1{)}^n$ survives, and with eight odd and seven even indices ${\sum }_{n=1}^{15}(-1{)}^n=-1$, so the sum equals $-2$ and its modulus is $2$.