Let ω=e^2π i/7 and S=ω+ω^2+ω^4.

Let $\bar S=\omega^6+\omega^5+\omega^3$ be the conjugate. Then $S+\bar S=\omega+\omega^2+\dots+\omega^6=-1$. In the product $S\bar S$ the three diagonal terms give $3$, and the six cross terms are $\omega^{k}$ for each $k=1,\dots,6$ exactly once (the differences of $\{1,2,4\}$ form $\{1,\dots,6\}$ once each), summing to $-1$; so $S\bar S=3-1=2$. Hence $S$ satisfies $z^2+z+2=0$, giving $S=\frac{-1\pm i\sqrt7}{2}$; numerically $S\approx-0.5+1.3229i$.
(A) True: $\operatorname{Re}S=-\frac12$.
(B) False: $|S|^2=2$, so $|S|=\sqrt2$, not $2$.
(C) True: $S^2+S+2=0$.
(D) False: $S^3=2-S\ne1$ (using $S^2=-S-2$).