Let $z$ be a complex number with $|z|<1$. For every subset $A\subset\mathbb N$, the series $\sum\_{m\in A}z^m$ is convergent. Denote $S(A)\in\mathbb{C}$ its sum and $\Sigma\_z$ the set of all numbers $S(A)$. Remark that the cardinal of $\Sigma\_z$ is (likely) that of ${\cal P}({\mathbb N})$, the continuum.



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> Is it possible that $\Sigma\_z$ be a neighbourhood of the origin ?
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Partial sums of $\sum_0^\infty z^n$
The number $z = i/\sqrt2$ seems to work!


Given $x \in [-2/3,4/3]$ we can find a "negabinary" expansion
$$x = \sum\_{k=0}^\infty (-1)^k\frac{b\_k}{2^k},$$ where each $b\_k \in \{0,1\}$.
Similarly, given $y \in [-2/3\sqrt2,4/3\sqrt2]$ we can find
$$y = \frac{1}{\sqrt2}\sum\_{j=0}^\infty (-1)^j\frac{c\_j}{2^j},$$
where each $c\_j \in \{0,1\}$.
Therefore,
$x + iy = \sum\_{n \in A} z^n$ where
$$A = \{2k : b\_k = 1\} \cup \{2j+1 : c\_j = 1 \}.$$




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As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $|z| \geq 1/\sqrt2$ is necessary. By definition, $$\Sigma\_z = z\Sigma\_z\cup(1+z\Sigma\_z).$$ 
If $\mu$ denotes Lebesgue measure, then $\mu(z\Sigma\_z) = |z|^2\mu(\Sigma\_z)$ thus $\mu(\Sigma\_z) \leq 2|z|^2\mu(\Sigma\_z)$. Since $\Sigma\_z$ is compact, it has finite measure and thus if $\mu(\Sigma\_z)>0$ then we must have $|z|^2 \geq 1/2$.


It remains open whether $|z|\geq1/\sqrt2$ and $z \notin \mathbb{R}$ is sufficient for $\Sigma\_z$ to contain $0$ in its interior.


