For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$ For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$
My solution :
let $$p = \alpha + \alpha^2 + \alpha^4$$ and $$q = \alpha^3 + \alpha^5 + \alpha^6.$$
We know $$1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 = 0,$$
$$p + q = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$$
$$ = -1$$
and
$$pq = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6)$$
$$= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^7 + \alpha^9 + \alpha^{10}$$
$$= 2.$$
Therefore, $p$ and $q$ are the two roots of the following equation :
$$x^2 + x + 2 = 0$$
and
$$p = \alpha + \alpha^2 + \alpha^4$$
$$= \frac{-1 ± \sqrt{7} i}{2}.$$
Would there be other ways of evaluating? I'm thinking of polar forms but not sure how to do this with it.

Let $p=\alpha+\alpha^2+\alpha^4$. Squaring both sides we obtain
$$p^2= \alpha^2+\alpha^4+\alpha^8+2\alpha^3+2\alpha^6+2\alpha^5.$$ Re arrange the terms we get $$p^2= \alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha^3+\alpha^5+\alpha^6.$$
Which is equal to $p^2=-1+ \alpha^3+\alpha^5+\alpha^6$ same as $p^2+1=\alpha^3+\alpha^5+\alpha^6$. Again squaring both sides  and re arrange the terms we obtain
$(p^2+1)^2=(\alpha^3+\alpha^5+\alpha^6)^2.$ $(p^2+1)^2=\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6+\alpha+\alpha^2+\alpha^4$. Which is same as $(p^2+1)^2=-1+p \Rightarrow p^4+2p^2-p+2=0.$ The root of this equation gives the desired result.
