For $S \subset \{z \in \mathbb{C}: z^{11}=1\}$ we define $z_s= \sum_{s \in S}s$.
Let $\zeta \in \mathbb{C}$ be a primitive $11$th root of unity. Is $z_s \in \mathcal{C}(0,1)$, for $S=\{1,\zeta,\zeta^3,\zeta^4,\zeta^5,\zeta^9\}$?
For how many subsets $S$ of $ \{z \in \mathbb{C}: z^{11}=1\}$ is $z_s \in \mathcal{C}(0,1)$?
$z_s$ is constructible iff $[\mathbb Q(z_s):\mathbb Q]$ is a power of $2$. The field extension $\mathbb Q(\zeta)/\mathbb Q$ is Galois, with the Galois group $(\mathbb Z/11\mathbb Z)^\times$, which is isomorphic to $C_{10}$ (cyclic group with $10$ elements). Since $C_{10}\cong C_5\times C_2$, there are only $4$ subgroups in $C_{10}$: $C_{10}$, $C_5$, $C_2$, $1$. In particular, $C_5$ is the subgroup of squares in $C_{10}$; in $(\mathbb Z/11\mathbb Z)^\times$ this subgroup is $\{1,3,4,5,9\}$ (non-zero squares modulo $11$).
By Galois correspondence, subgroups give us subfields, and the index of the subgroup is the degree of the subfield over $\mathbb Q$. In other words $z_s$ is constructible iff the index of $G$ in $(\mathbb Z/11\mathbb Z)^\times\cong C_{10}$ is a power of $2$, where $G$ is the subgroup fixing $z_s$. So either $G=C_5$ (index $2$) or $G=C_{10}$ (index 1, in this case $z_s$ is rational). In other words, $z_s$ is constructible iff it's fixed by $C_5$.
Now $z_S$ is $C_5$-invariant iff $S$ is $C_5$-invariant (this requires a bit of thought: there is one linear relation between the powers of $\zeta$, namely $1+\zeta+\dots +\zeta^{10}=0$, but it will not spoil this result). So the possible $S$'s are: $\{\zeta,\zeta^3,\zeta^4,\zeta^5,\zeta^9\}$, $\{\xi,\xi^3,\xi^4,\xi^5,\xi^9\}$ where $\xi=\zeta^2$, $\{1\}$ (these are the 3 $C_5$-orbits), and their unions.
(For more info look up Gauss periods and Gauss sums.)