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.

Is it possible that $\Sigma_z$ be a neighbourhood of the origin ?

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    $\begingroup$ The question is: Does there exist $|z|<1$ such that $\Sigma_z=\{\sum\limits_{n\in A} z^n: A\subseteq \mathbb N_0\}$ is a neighbourhood of $0$. $\endgroup$ – Jochen Wengenroth May 27 at 8:00
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    $\begingroup$ @JochenWengenroth Why is $\alpha\notin\mathbb Q$ necessary when $r<1$? $\endgroup$ – Emil Jeřábek May 27 at 9:14
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    $\begingroup$ To record what I just understood from @JochenWengenroth's comment: $r \geq 1/2$ is necessary because $\Sigma = z\Sigma \cup (1 + z\Sigma)$ and therefore $\mu(\Sigma) \leq 2r\mu(\Sigma)$ where $\mu$ is Lebesgue measure. ($\Sigma$ is evidently compact and therefore of finite measure.) $\endgroup$ – François G. Dorais May 27 at 9:28
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    $\begingroup$ I agree with @EmilJeřábek. Wouldn't $z = i/\sqrt{2}$ work knowing that every integer has a negabinary representation? mathworld.wolfram.com/Negabinary.html $\endgroup$ – François G. Dorais May 27 at 9:43
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    $\begingroup$ @FrançoisG.Dorais $\mu(z\Sigma) = r^2 \mu(\Sigma)$, right? So that gives $r \geq 1/\sqrt 2$. $\endgroup$ – Bart Michels May 27 at 12:18

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 $$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$

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.

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  • $\begingroup$ I'm a bit tired so the interval bounds might be a bit off... Please check my arithmetic! $\endgroup$ – François G. Dorais May 27 at 10:10
  • $\begingroup$ Denis: thank you for correcting my accidental abuse of the letter $i$. For what its worth, I think the case $z= i/\sqrt2$ is anecdotal and doesn't fully answer your question. $\endgroup$ – François G. Dorais May 27 at 11:02
  • $\begingroup$ François, it answers my question. Of course, it raises the more general question of which z has this property ?. By the way, how do you prove that $\Sigma_z$ is measurable ? $\endgroup$ – Denis Serre May 27 at 13:38
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    $\begingroup$ Denis, $\Sigma_z$ is compact since it is a continuous image of $2^{\mathbb N}$. Therefore it is measurable and has finite measure. $\endgroup$ – François G. Dorais May 27 at 16:05
  • $\begingroup$ See Davis, Chandler; Knuth, Donald E. Number representations and dragon curves-I. J. Recreational Math. 3 (1970), no. 2, 66–81. and part II in J. Recreational Math. 3 (1970), no. 3, 133–149. $\endgroup$ – Gerald Edgar May 27 at 16:26

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