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.
 A: 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.
A: The extension $\mathbb Q(\alpha)/\mathbb Q$ has Galois group $(\mathbb Z/7\mathbb Z)^\times\cong\mathbb Z/6\mathbb Z$, generated by $3\in(\mathbb Z/7\mathbb Z)^\times$. Denote the image of $a$ under the isomorphism $(\mathbb Z/7\mathbb Z)^\times\to\mathrm{Gal}(\mathbb Q(\alpha)/\mathbb Q):a\mapsto (\alpha\mapsto\alpha^a)$ as $\sigma_a$.
Now, since $u=\alpha+\alpha^2+\alpha^4$ is invariant under $\sigma_2$, it must be in the fixed field of $\langle\sigma_2\rangle$, which has index $2$ in $\mathrm{Gal}(\mathbb Q(\alpha)/\mathbb Q)$. Thus, under the Galois correspondence, it must be contained in the degree-$2$ extension $\mathbb Q(\sqrt{-7})$. Since $u$ is integral, it is contained in $\mathbb Z\big[\frac{1+\sqrt{-7}}2\big]$.
Since $\sigma_3$ is the generator of the Galois group of $\mathbb Q(\sqrt{-7})/\mathbb Q$, it is complex conjugation. We have $u+\sigma_3(u)=\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6=-1$, so $u$ has real part $-\frac12$. We conclude $u=-\frac12+\frac{2n-1}2{\sqrt{-7}}$ for some integer $n$.
We are almost done at this point, and we have several ways to finish off:

*

*Look at the absolute values. $|-\frac12+\frac{2n-1}2{\sqrt{-7}}|^2=\frac14+\frac{(2n-1)^2}4\cdot7$, while $|u|^2\le(|\alpha|+|\alpha|^2+|\alpha|^4)^2=9$. We thus need $n=0,1$, giving the desired result.


*Calculate the norm of $u\in\mathbb Q(\sqrt{-7})$ which is $u\cdot\sigma_3(u)$, which turns out to be $2$. Comparing with the formula $u=-\frac12+\frac{2n-1}2\sqrt{-7}$, we again obtain $n=0,1$.
A: 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 $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.
