Is every Pisot-like integer the product of a Pisot integer and a root of unity? For lack of better terminology, let's call an algebraic integer $\beta$ Pisot-like if $|\beta|_{\mathbf{v}} > 1$ for the place $\mathbf{v}$ of $\Bbb{Q}(\beta)$ corresponding to the embedding $\beta \mapsto \beta$ and $|\beta|_v < 1$ for every other infinite place of $\Bbb{Q}(\beta)$.
With this in mind, a Pisot integer (or Pisot number, if you prefer) is a Pisot-like positive real algebraic integer.
Question: Is every Pisot-like integer the product of a Pisot integer and a root of unity?

What I have so far:
Clearly, if $\beta$ is Pisot-like and $\zeta$ is a root of unity, then $\zeta\beta$ is Pisot-like because $|\cdot|_v$ is multiplicative for every place $v$ of $\Bbb{Q}(\beta)$.
On the other hand, for any Pisot-like number we can find a rotation of the complex plane which sends it to a real positive Pisot-like number. This would be enough to conclude if I knew, for example, that every rotation of the complex plane which preserves integrality can be written as the product by a root of unity.
 A: Here's a counterexample. Consider $\beta \in \Bbb{Z}[i] \setminus \Bbb{Z}$. Then $\Bbb{Q}(\beta) = \Bbb{Q}(i)$ has exactly one archimedean place, because it is a purely complex extension of degree $2$ over $\Bbb{Q}$. Hence $\beta$ is Pisot-like iff $|\beta| > 1$, where $|\cdot|$ is the usual absolute value on $\Bbb{C}$, which happens iff $\beta \neq \pm i$.
Finally, note that $|\beta|$ is always an alegbraic integer; if it is not Pisot, then $\beta$ cannot be obtained from a Pisot integer through any rotation of the complex plane. One such $\beta$ is $1+i$.
A: There are counterexamples of higher degrees as well. I post my argument, because some of the ideas may be reusable.
Consider the polynomial
$$
q(x)=(x^2-4x-1)^2+1.
$$
By Eisenstein's criterion ($p=2$) $q(x)$ is irreducible.
Its zeros are zeros of either of $x^2-4x-(1\pm i)$, so they are
$$
u_1=2+\sqrt{5+i},\ u_2=2+\sqrt{5-i},\ u_3=2-\sqrt{5+i},\ u_4=2-\sqrt{5-i}.
$$
Let $L=\Bbb{Q}(\sqrt{5+i},\sqrt{5-i})$ be the splitting field of $q(x)$. Because $5+i=(1+i)(3-2i)$ (resp. $5-i=(1-i)(3+2i)$ it is easy to show that the extension $\Bbb{Q}(\sqrt{5+i})/\Bbb{Q}(i)$ ramifies above the prime $3+2i$ (resp. the extension $\Bbb{Q}(\sqrt{5-i})/\Bbb{Q}(i)$ ramifies above the prime $3-2i$) but does not ramify above the conjugate prime, so these two extensions are distinct. Therefore $[L:\Bbb{Q}]=8$.
Let us view the Galois group $G=Gal(L/\Bbb{Q})$ as a subgroup of $S_4$ (use the subscripts of the roots). We see that there are automorphisms $\sigma_1=(24)$ and $\sigma_2=(13)$ in $G$, because these are the automorphism of $L$ over $\Bbb{Q}(i)$. Furthermore, the complex conjugation acts as $\sigma_3=(12)(34)$. The group they generate is a copy of the dihedral group $D_4$ of eight elements.
For the purposes of the problem at hand we specify that $\sqrt{5\pm i}$ are the complex numbers with arguments $\pm\dfrac12\arctan\dfrac15$. This means that
$|u_1|=|u_2|>4$ and $|u_3|=|u_4|<1$, so we have a Pisot situation.
I claim that no power of $u_1^k$, $k$ a positive integer, is a real number. This is equivalent to saying that $u_1$ cannot be gotten from a real algebraic integer by multiplying it with a root of unity.
One of the keys comes from Galois theory. The subfield $L\cap\Bbb{R}$ corresponds to the group $\langle \sigma_3\rangle\le G$. OTOH the subfield
$\Bbb{Q}(u_1)$ corresponds to the group $\langle \sigma_1\rangle\le G$. But the subgroup corresponding to $\Bbb{Q}(u_1)\cap \Bbb{R}$ is
$$
\langle\sigma_1,\sigma_3\rangle=\langle(24),(12)(34)\rangle=D_4
$$
all of $G$, so we get that $\Bbb{Q}(u_1)\cap\Bbb{R}=\Bbb{Q}$. So if $u_1^k$ is a real number for some $k$, then $u_1^k$ needs to be rational (because it is in that intersection of fields). Furthermore, $u_1^k$ is an algebraic integer, so it needs to be a rational integer as well.
But if $u_1^k=m\in\Bbb{Z}$, then an application of a suitable element of $G$ implies that $u_3^k=m$ also. But $0<|u_3|^k<1$, so this is absurd.
