Is there a way to see if $\alpha \in \mathbb{C}$ is constructible at a glance? The notion of constructibility is not too obscure but mathematically, I find the definitions tedious and not very easy to handle with.
I don't know if Ian Stewart's book Galois Theory edition 4 thinks the same, but it mostly glosses over with the reasoning of whether some $\alpha \in \mathbb{C}$ s constructible or not.
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There exists an angle that cannot be trisected. Let's prove this by giving a specific example of such an angle; we look at $\frac{2 \pi}{3}$.

See, then it starts off by saying

Let $\omega=e^{\frac{2 \pi}{3}i}$. Then $\omega$ is constructible since $\omega=\frac{-1+i\sqrt{3}}{2}$.

$\omega$ is constructible since $\omega=\frac{-1+i\sqrt{3}}{2}$? How did he conclude that in a fraction of a second? 
Is there an unspoken loose rule? Perhaps it's possible to rigourously, concretely prove that $\omega$ is constructible, but that's not what I am looking for. It seems like these people can see some complex number for a few seconds and say "that's constructible" or "nah that doesn't look constructible"
How can you tell? If it's expressible in fractions? If it's a number along the unit circle? But $\omega$ has an irrational in it, and does that matter? Does it not matter?
I can't get a grip on this "constructible numbers" thing, especially telling whether or not it is! If someone can provide me a general argument in judging/guesstimating the constructibility of a particular complex, that would be great. How do you guys do it?
 A: $\frac{-1+i\sqrt 3}{2}$ is constructible because we only need to adjoin square roots to obtain that number. Specifically, we only need $i$ (the square root of $-1$) and $\sqrt 3$ (the square root of, well, $3$). In fact, the product $i\sqrt 3$ is by itself a square root of the rational ("obviously constructible" would suffice) number $-3$. The rest is just field operations. 
A: A complex number $z$ is constructible if there is a series $$\Bbb Q \subset K_1, \quad K_1 \subset K_2, \quad \ldots, \quad K_{n - 1} \subset K_n$$
of quadratic field extensions such that $z$ is in the last field, $K_n$.
In particular, $$\omega = e^{2 \pi i / 3} = -\tfrac{1}{2} + \tfrac{1}{2} \cdot i \sqrt{3}$$ is in $\Bbb Q(i \sqrt{3})$, but $i \sqrt{3}$ is a root of the quadratic rational polynomial $x^2 + 3$, so the extension $\Bbb Q \subset \Bbb Q(i \sqrt{3})$ is quadratic and hence $\omega$ is constructible.
Alternatively, $\omega$ satisfies $\omega^3 = 1$. Rearranging this gives $(\omega - 1)(\omega^2 + \omega + 1) = 0$, and since $\omega \neq 1$ $$\omega^2 + \omega + 1 = 0 .$$ Thus, $\Bbb Q(\omega) \ni \omega$ is a quadratic extension of $\Bbb Q$.
