Non real complex in metric completions of $\mathbb Q$ Process of completion of $\mathbb Q$ using the absolute value $|x|$ does not touch to the non-real complex numbers which are added to $\mathbb Q$ via extensions fields. However completion of $\mathbb Q$ using another kind of absolute value can aggregate complex numbers to $\mathbb Q$. (The process is formally the same: the ring $R$ of Cauchy sequences of $\mathbb Q$, in which addition and multiplication are defined termwise as usually, has the maximal ideal $M=\{(r_n)_{n\in\mathbb N}\in R:  r_n \to 0\}$ so $R/M$ is a field, which gives a definition of $\mathbb R$ when$|x|=\max\{x,-x\}$, the Euclidean absolute value).
According to the sometimes called Ostrowski’s Theorem, all the (not trivial) absolute values on $\mathbb Q$ (except equivalent ones) are only the p-adic absolute values (if the non-zero rational $x$ is written as $x= p^n \frac{a}{b}$,with $a,b,p$ pairwise coprime and $n$ rational integer, its definition is $|x|_p = p^{-n} $ and $|0|_p=0$).
The question is about the following known result: “All the $(p-1)$-th roots of $1$ are elements of $\mathbb Q_p$” (it can be proved using residue field of $\mathbb Q_p$ as isomorphic to $\mathbb F_p$ and applying Hensel’s lemma to the polynomial $x^{p-1}-1$ of $\mathbb F_p[x]$)
As much as I try I fail to give me an entirely satisfactory understanding about why the completions of  $\mathbb Q$ via $p$-adic absolute values  contain a lot of non-real complex and the ordinary completion of $\mathbb Q$ does not contain any of them. I know there are quite "pathology" in $\mathbb Q_p$ but topological and metric, not of “inclusion of elements”. I do not even know by now how to write the $p$-adic expansion of $e^\frac {2\pi i k}{n}$ for adequate $k$ and $n$. Some help?
 A: When you’re speaking of the $p$-adic domain, the concepts of “real” and “complex” don’t make sense. Those concepts relate only to the archimedean metric on $\Bbb Q$.
You’re correctly observing that every root of the $n$-th cyclotomic polynomial is complex, but what about $X^6-5$? Irreducible, all the roots are equivalent in a Galois-theoretic viewpoint, and if you look at the abstract field $\Bbb Q(\xi)/(\xi^6-5)$, is $\xi$ real or complex? The right way to look at this is that the abstract field has two inequivalent real absolute values, and four complex, though the latter fall into equivalent pairs. If you know algebraic number theory, $r_1=r_2=2$ here.
As to your question about $e^{i\pi/n}$, this is doubly meaningless in the $p$-adic domain. First, there’s no $\pi$ there; and second, although there is an exponential function defined on a relatively small open neighborhood of $0$, there’s no reason to think that your “$i\pi/n$” would lie in it.
I’m sorry for this diatribe. Lang told us, when I was in college, that the $p$-adic domains were just as good as the real and complex domains.
Here’s something that is a wonderful tool $p$-adically, but has no equivalent in the complex domain. It’s the $p$-adic logarithm, which is defined for all $\xi$ with $|\xi-1|<1$. This set, a pretty large neighborhood of the multiplicative identity $1$, is already a group (!!)
in the $p$-adic domain, even when you make algebraic extensions to $\Bbb Q_p$. And the logarithm, defined by the very same power series that you learned in Calculus, is a homomorphism (!!) defined on that group. This is something that’s not true in the complex domain: the logarithm, in whichever way you define it on an open subset of $\Bbb C$, is not a homomorphism, because it can’t be defined on an open multiplicative subgroup of $\Bbb C$.
