Let $z = e^{ \frac {2\pi i}{7}} $ and let $p = z+z^2+z^4 $.Then which of the following options are correct?

Question

Let $z= e^{ \frac {2\pi i}{7}} $ and let $p= z+z^2+z^4 $ then

1. $p$ is in $ \mathbb {Q} $

2. $p$ is in $ \mathbb{Q} (\sqrt {D}) $ for some $D \gt 0$

3. $p$ is in $ \mathbb{Q}(\sqrt {D}) $ for some $D \lt 0$

4. $p$ is in $i \mathbb {R} $

Option $1$ is clearly false. please give me some hints for other options.

Thanks in advance.

Answer 1

Note that $$p^2=z^2+z^4+z+2z^3+2z^5+2z^6=-p-2+2(1+z+z^2+z^3+z^4+z^5+z^6)$$ and $1+z+\cdots+z^6=0$.

Answer 2

Since $p^\ast=z^6+z^5+z^3=z^2p$ and $z^2\ne\pm1$, $p$ is neither real nor imaginary. eliminating $1$, $2$ and $4$. As Lord Shark the Unknown already noted, $p^2+p+2\implies p=\frac{-1\pm\sqrt{-7}}{2}$ so option $3$ is correct.