7th roots of unity

Question

I'm having difficulty expressing 7th roots of unity with nth root. Here in the photo are my solutions, but they are too complicated and I'm not sure if they are correct. If you would tell me if they are correct, or how to solve the problem, that would be very helpful. Thank you.

Answer 1

You cannot express the $7$th roots of unity in the form $a+bi$ where $a$ and $b$ are rendered with rational numbers and real radicals.

What I mean is demonstrated by contrasting with the fifth roorts of unity which have the form

$\dfrac{-1\pm_1\sqrt5}4\pm_2i\sqrt{\dfrac{5-(\pm_1\sqrt5)}8}$

(where the $\pm$ signs with the same subscript are to be chosen identically). Note that the real and imaginary parts are separately made up of real radicals; the only non-real radical is the one implied by the factor $i$ in the second term. But an expression with these properties does not exist for $7$th roots of unity or any other $p$-root of unity except where $p$ is (1) a power of $2$, (2) a square-free product of Fermat primes, or (3) some product of (1) and (2).

What blocks you is the casus irreducibilis, a mathematical phenomenon that emerges when you try to express multiple real roots of an irreducible odd-degree polynomial with real quantities only. It is rigorously proven that you can't. In the specific case of primitive seventh roots of unity, it is easy enough to show that twice the real parts of these primitive roots solve the cubic equation $x^3+x^2-2x-1=0$ and that this equation properly has three real roots for the required three complex-conjugate pairs. But with the casus irreducibilis, those three real roots cannot be expressed entirely with rational numbers and real radicals.