Let $z\_i \in \mathbb{C}\:$ for $i=1,\dots, n\;$ be complex numbers, all with absolute value $|z\_i|\le 1\;$.


Prove (or disprove) that there exists a choice of signs $s\_i \in \{\pm 1\}$ such that
$$\left|\sum\_{i=1}^n s\_i\cdot z\_i\right| \le \sqrt{2}.$$


[My interest in this problem is purely for fun. I couldn't solve it a long time ago, forgot about it, but shortly ago it came back into my mind again.]



Bounding a signed sum of complex numbers
What follows is not an answer, but is too long for a comment.


This problem and its natural higher-dimensional generalization is connected with the recent MO questions [Covering a unit ball with balls half the radius](https://mathoverflow.net/questions/98007/covering-a-unit-ball-with-balls-half-the-radius) and [covering disks with smaller disks](https://mathoverflow.net/questions/97939/) : 
let $K\_d$ be the smallest constant such that for any sequence $(z\_i)\_{i \geq 1}$ of vectors of $\mathbb{R}^d$ of (euclidean) norm at most one, there's some choice of signs $s\_i = \pm 1$ such that the partial sums $\sum\_{1 \leq i \leq n} s\_i z\_i$ are all bounded by $K\_d$.


Now let $N\_d$ be the minimal number of balls of radius $\frac{1}{2}$ needed to cover a ball of radius $1$ (in $\mathbb{R^d}$). I claim that $K\_d \leq N\_d$.


Proof : Let $K\_{d,n}$ be the same constant as $K\_d$, but for which we require only the first $n$ partial sums to be bounded by $K\_{d,n}$. Then a straightforward averaging argument yields $K\_{d,n} \leq \sqrt{n} \leq n$. Now let $n > N\_d$. Fixing a covering of the unit ball with $N\_d$ balls of radius $\frac{1}{2}$, then there must be two distinct $ i < j \leq N\_d +1$ such that $z\_i$ and $z\_j$ lie in the same ball of radius $\frac{1}{2}$, and hence must satisfy $|| z\_i - z\_j || \leq 1$. If we replace $z\_j$ by $z\_j - z\_i$, suppress $z\_i$, and then use $K\_{d,n-1}$, we get a sequence of signs which achieve $K\_{d,n} \leq \max ( N\_d, K\_{d,n-1} ) $. But Kônig's lemma (for infinite binary trees) gives $K\_d \leq \sup\_{n} K\_{d,n} $, hence the desired result.


