Is it true that $\mathbb{Q}^r(i)$ contains any root of unity, when $\mathbb{Q}^r$ is real root closure of $\mathbb Q$?

Question

Define a real root extension of $\mathbb{Q}$ to be a sequence of extensions of the form $$ \mathbb{Q}\subset \mathbb{Q}(a_1) \subset \mathbb{Q}(a_1, a_2) \subset \ldots \subset \mathbb{Q}(a_1, \ldots, a_r)\subset \mathbb{R} $$ where a power of $a_1$ lies in $\mathbb{Q}$ and for each $i\in\{2,\ldots r\}$, a power of $a_i$ lies in $\mathbb{Q}(a_1, \ldots, a_{i-1})$.

Let $\mathbb{Q}^r$ be the union of all possible real root extensions of $\mathbb{Q}$. Is it true that $\mathbb{Q}^r(i)$ contains any root of unity?

Answer 1

The primitive root of unity $\cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$ belongs to $\mathbb{Q}^r(i)$ if and only if $\varphi(n)$ is a power of $2$, where $\varphi(n)$ is the Euler's totient function. Such numbers $n$ are of the form $2^kp_1p_2\ldots p_s$, where $p_i$ are distinct Fermat primes.

This follows from the following fact: $\cos(\frac{2\pi}{n})$ and $\sin(\frac{2\pi}{n})$ can be written using only rational numbers, addition, subtraction, multiplication, division, and roots of positive numbers if and only if $\varphi(n)$ is a power of $2$.

See nice exposition in: Skip Garibaldi, Somewhat more than governors need to know about trigonometry, Mathematics Magazine 81 (2008) #3, 191-200.