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.

Yet another way (albeit not too different from OP's), using that $\,|\alpha| = 1\,$, $\,\overline \alpha=\dfrac{1}{\alpha}\,$:

*

*$\require{cancel}\displaystyle \;p+\overline p = \alpha + \alpha^2+\alpha^4 + \overline \alpha + \overline \alpha^2 + \overline \alpha^4 =\alpha + \alpha^2+\alpha^4 + \frac{1}{\alpha}+\frac{1}{\alpha^2}+\frac{1}{\alpha^4} \\\displaystyle =\frac{\cancel{\alpha^5+\alpha^6+\alpha+\alpha^3+\alpha^2+1\color{red}{+\alpha^4}}-\color{red}{\alpha^4}}{\alpha^4}=\frac{-\alpha^4}{\alpha^4}=-1$


*$\displaystyle \;|p|^2=\cancel{|\alpha|^2}\,|1+\alpha+\alpha^3|^2=\left(1+\alpha+\alpha^3\right)\left(1+\overline\alpha+\overline\alpha^3\right)= \left(1+\alpha+\alpha^3\right)\left(1+\frac{1}{\alpha}+\frac{1}{\alpha^3}\right) \\\displaystyle = \frac{\left(1+\alpha+\alpha^3\right)\left(1+\alpha^2+\alpha^3\right)}{\alpha^3}=\frac{\cancel{1+\alpha^2+\alpha^3+\alpha}+\alpha^3+\cancel{\alpha^4}+\alpha^3+\cancel{\alpha^5+\alpha^6}}{\alpha^3}=2$
The two relations give the real part $\dfrac{-1}{2}$ and magnitude $\sqrt{2}$ of $\,p\,$, so $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.
