<s>[INST] 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.
 [/INST] 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 $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.
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