Prove that exist $e_1,\dots,e_n\in\{-1,1\}$ such that $|e_1z_1+{\dots}+e_nz_n|\le\sqrt2$

Let $z_1,\dots,z_n\in\mathbb{C}$ such that $|z_p|\le1$ for every $p\in\{1,\dots,n\}$. Prove that exist $e_1,\dots,e_n\in\{-1,1\}$ such that $|e_1z_1+{\dots}+e_nz_n|\le\sqrt2$.

I have firstly tried using induction. For $n=1$ I have $|e_1z_1|=|z_1|\le1\le2$, so it is true. Then I supposed that $|L|\le\sqrt2$ where $L=e_1z_1+{\dots}+e_kz_k$ for some $k\in\mathbb{N},k\le n$. Then I tried to prove that it is true for $k+1$. But then I realized that it is impossible for any $z_{k+1}$ if I do not change values of $e_1,\dots,e_k$ because $|z_p|\le1$, but $|L|\le\sqrt2$. My second attempt is to write numbers $z_1,\dots,z_n$ in trigonometric form. Let $|z_p|=r_p$ and $\arg(z_p)=a_p$. Then I need to prove that $$\sqrt{(e_1r_1\cos a_1+{\dots}+e_nr_n\cos a_n)^2+(e_1r_1\sin b_1+{\dots}+e_nr_n\sin b_n)^2}\le\sqrt2$$ but I do not know how it can help me to prove main inequality. What is the easiest way to prove it?

Solution 1:

Partial Result:

First divide the unit disk into equal sectors as shown below.

Notice that adding two vectors from opposing sectors results in a vector also in the unit circle. Since we can choose whether a vector is in a sector or its opposite (by possibly multiplying by $-1$), we can reduce the problem to the case $n=3$ (where each vector lies in a different colored area).

Solution 2:

Peter Woolfitt's answer basically solves the problem. As he explains, we can reduce to the case $n=3$, where $\arg(z_1)$, $\arg(z_2)$ and $\arg(z_3)$ lie in $[0, \pi/3]$, $[2 \pi/3, \pi]$ and $[4 \pi/3,5 \pi/3]$ respectively. But, if $w$ and $z$ are in the unit disc and $2 \pi/3 \leq \arg(w) - \arg(z) \leq \pi$, then $w+z$ is in the unit circle. This will apply for one of the three pairs $(z_1, z_2)$, $(z_2, z_3)$ and $(z_3, z_1)$ so we reduce to $n=2$.

Possibly replacing $z_2$ by $-z_2$, we can assume that $z_1$ and $z_2$ make an obtuse angle with each other. Then $|z_1+z_2| \leq \sqrt{2}$, as desired.