This is somewhat nagging me. I just with to run it by some one cause it seems slightly unintuitive to me (in some regards).
Let us consider $r$ to be an algebraic number in $\mathbb{C}$ not in $\mathbb{R}$.
Clearly it follows then that $\bar{r}$ is also an algebraic number.
As a consequence then $\frac{r}{|r|}$ is an algebraic number. Note also that $\frac{r}{|r|}$ lies on the unit circle in $\mathbb{C}$.
Let us call $\frac{r}{|r|} = u = e^{i \cdot \theta}$
Note then that $\mathbb{Q}(u)$ is a finite extension of $\mathbb{Q}$
Also please note that if $\theta$ is not a rational multiple of $\pi$ then the set of numbers:
$\{ u^k : 0 \le k \le \infty \}$ must be dense over $S_1$ the unit circle in $\mathbb{C}$
The latter clearly implies that if $\theta$ is not a rational multiple of $\pi$ then it must be the case that the extention $\mathbb{Q}(u)$ is an infinite extension. Which directly contradicts the given we started with.
So the conclusion is that if $r$ is any algebraic number in $\mathbb{C}$ then the fraction $\frac{r}{|r|}$ is a root of unity.
As a consequnce then we have that $u$ is an element of $\mathbb{Q}^{ab}$ or the abelian closure of $\mathbb{Q}$.
If my memory of Galois Theory serves all polynomials with roots in $\mathbb{Q}^{ab}$ can be solved algebraically as nested radicals.
Sorry this seems a little odd ... perhaps it may be true.
This step is incorrect. You have not justified this "clearly" and it is just not true (I don't know what argument you have in mind but $\mathbb{Q}[u]$ can be dense in $\mathbb{C}$ as soon as its dimension is at least $2$). You can already find counterexamples in $\mathbb{Q}[i]$, which look like $\frac{a + bi}{a - bi}$ for $a, b \in \mathbb{Z}$, e.g. already $\frac{1 + 2i}{1 - 2i}$ is not a root of unity. Of course this is an abelian extension but counterexamples in nonabelian extensions are harder to work with. The point is that this step doesn't work.