JEE Main · 2006 · 30 Apr 2006 · Offline · Complex Numbers

The condition $z^2+z+1=0$ makes $z$ a primitive cube root of unity, so $z^3=1$, $1+z+z^2=0$ and $\frac{1}{z}=z^2$. Then $z+\frac{1}{z}=z+z^2=-1$ and likewise $z^2+\frac{1}{z^2}=z^2+z=-1$, while $z^3+\frac{1}{z^3}=1+1=2$. The powers repeat with period three, so the six quantities $z^k+\frac{1}{z^k}$ for $k=1,\dots ,6$ are $-1,-1,2,-1,-1,2$. Squaring each gives $1,1,4,1,1,4$, and their sum is $1+1+4+1+1+4=12$.