Is there a simple procedure to produce algebraic numbers of modulus one that are not roots of unity? Let $z=e^{i\theta}$ be a complex number of modulus $1$. Trivially
if $z$ is a root of unity then $z$ is also an algebraic number, but the converse
is known to be false : $z$ can be algebraic without being a root of unity (consider, e.g. $\frac{3+4i}{5}$).
Is there a simple procedure to produce for each $n$, an algebraic number
$z_n$ in this situation, with degree $n$ over ${\mathbb Q}$ ?
There are several related questions on this theme on MSE, for example 
here. 
 A: The comment by Sea Turtles shows that we need to assume that $n$ is even, say $n=2k$. I proffer a general recipe that works for all $k\ge3$. To add something extra to the recipe in Qiaochu Yuan's comment I add the property that $z_n$ should be an algebraic integer to the wish list.
Let $\alpha=\root k\of 2 -1$. The minimal polynomial of $\alpha$ is $(x+1)^k-2$ which is irreducible by Eisenstein, so $[\Bbb{Q}(\alpha):\Bbb{Q}]=k$. We have clearly $0<\alpha<1$,
so if $\beta=\sqrt{1-\alpha^2}$, then $i\beta$ is purely imaginary, and $|\alpha+i\beta|=1$.
I claim that $z_n=\alpha+i\beta$ works.
Because $i\beta$ and $\alpha$ are algebraic integers, so is $z_n$. We also have $[\Bbb{Q}(\alpha,i\beta):\Bbb{Q}(\alpha)]=2$. Furthermore,
$$
z_n+z_n^{-1}=2\alpha,\qquad\text{and}\qquad z_n-z_n^{-1}=2i\beta,
$$
so $\Bbb{Q}(z_n)=\Bbb{Q}(\alpha,i\beta)$. Thus $[\Bbb{Q}(z_n):\Bbb{Q}]=2k=n$.
It remains to show that $z_n$ is not a root of unity. Assume contrariwise that this would
be the case. By the well known Galois theory of cyclotomic fields this implies that the extension $\Bbb{Q}(z_n)/\Bbb{Q}$ is abelian and Galois. By basic Galois theory the same would hod for the subextension $\Bbb{Q}(\alpha)/\Bbb{Q}$.
But $\Bbb{Q}(\alpha)=\Bbb{Q}(\root k\of 2)$ is not normal, because the irreducible  polynomial $p(x)=x^k-2$ has some but not all of its roots in there.
