For what translation groups $G$ of $\mathbb C$ is $\mathbb C / G$ compact? If $G$ is a cyclic translation group (generated by $z \mapsto z + \tau$ for some $\tau \in \mathbb C$), then obviously $\mathbb C/G$ is not compact, since if $\tau = 0$ then $\mathbb C/G \cong \mathbb C$, and if $\tau \neq 0$ then $\mathbb C / G \cong [0, 1) + i\mathbb R$, which is also not compact.
The converse is not true; if $G$ is generated by $z \mapsto z + 1$ and $z \mapsto z + \pi$, then $\mathbb C / G$ is again not compact, even though $G$ is not cyclic.
So when is it the case that $\mathbb C/G$ is compact? How can we characterize such translation groups $G$?
 A: We may as well ask the more general question "For which subgroups 
$G \subset (\mathbb{R}^n, +)$ is $\mathbb{R}^n / G$ compact?".
The answer is that it is compact if and only if the linear span of $G$ is
all of $\mathbb{R}^n$.
If $G$ spans $\mathbb{R}^n$, it contains a basis, so we can find a linear 
isomorphism $f: \mathbb{R}^n \to \mathbb{R}^n$ such that 
$\mathbb{Z}^n \subset f[G]$. Then 
$\mathbb{R}^n/G \cong (\mathbb{R}^n/\mathbb{Z}^n) / (f[G]/\mathbb{Z}^n)$, which must be compact because 
$\mathbb{R}^n/\mathbb{Z}^n \cong \mathbb{T}^n$ is already compact.
If $G$ does not span $\mathbb{R}^n$, there is a linear isomorphism $h: \mathbb{R}^n \to \mathbb{R} \times\mathbb{R}^{n-1}$ such that 
$h[G] \subset \{0\} \times \mathbb{R}^{n-1}$. In that case, since $\mathbb{R}$ is locally compact, we have
$\mathbb{R}^n/G \cong \mathbb{R} \times (\{0\} \times \mathbb{R}^{n-1})/h[G]$, which is not compact because $\mathbb{R}$ is not.
Since $\mathbb{C}$ is a two-dimensional over $\mathbb{R}$, this means
that $\mathbb{C}/G$ is compact if and only if there are two elements
of $G$ that are linearly independent over $\mathbb{R}$, as suggested in
Blake's comments.
