For fixed $z_i$s inside the unit disc, can we always choose $a_i$s such that $\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3$?

Question

Let $z_1,z_2,\ldots,z_n$ be complex number such that $|z_i|<1$ for all $i=1,2,\ldots,n$. Show that we can choose $a_i \in\{-1,1\}$, $i=1,2,\ldots,n$ such that $$\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3.$$

Answer 1

I was not able to think it through properly, but here's a sketch:

Use induction as suggested by Berci, but with a little twist. The main idea is that for two numbers $z_i$ and $z_j$ such that $|z_i| < 1$ and $|z_j| < 1$ we can obtain $|z_i\pm z_j| < 1$ as long as some angle (out of four) between them (the difference of arguments) is smaller or equal than $\frac{\pi}{3}$. However, as long as we have 3 or more numbers, we will be able to find such a pair.

Quick illustration of the lemma: $z_i$ is somewhere on the blue line, red cross is the $z_j$ and the violet is their sum. The point is that as long as the red cross belongs to darker green, the violet line will stay in light green region.

$\hspace{70pt}$

I don't know if I will find enough time to work out all the details, so should this idea suit you, feel free to use it.

Cheers!

Answer 2

Claim: If $z_1, z_2, z_3, z_4$ are four numbers inside the open unit disk, then there is a pair of them $z_k, z_j$ with $z_k \pm z_j$ also in the unit disk, for the correct choice of sign.

Proof: If $z_1 = 0$, then $z_3 = z_3 + z_1$ and we're done. Otherwise, rotate the disk so that without loss of generality we can consider $z_1$ to be a positive real number. Let $b_i = \pm 1$ so that $b_2 z_2, b_3 z_3,$ and $b_4 z_4$ have non-negative imaginary part. Let $\theta_i = \arg(b_i z_i)$, with $0 \le \theta_i \le \pi$. Re-order the $z_i$ in terms of increasing argument so that $0 = \theta_1 \le \theta_2 \le \theta_3 \le \theta_4$. $\theta_4 = (\theta_2 - \theta_1) + (\theta_3 - \theta_2) + (\theta_4 - \theta_3) \le \pi $. There must be an index $j$ with $\theta_{j+1} - \theta_j$ no more than $\displaystyle \frac{\pi}{3}$. Let $w_1 = b_{j+1} z_{j+1}, w_2 = b_{j} z_{j}$. Then $e^{-i\theta_{j}}w_2$ is a positive real, and $0 \le \arg(e^{-i\theta_{j}}w_1) \le \displaystyle \frac{\pi}{3}$. It's easy to show that $$ |b_{j} z_{j} - b_{j+1} z_{j+1}| = |w_2- w_1| = |e^{-i\theta_{j}} w_2 - e^{-i\theta_{j}} w_1 | = |1 - e^{-i\theta_{j}}w_1| \lt 1 $$

But $|z_{j} \pm z_{j+1}| = |b_{j} z_{j} - b_{j+1} z_{j+1}|$ for one choice of sign, so we get the claim.

Now that we have the claim the rest is easy. Starting with any collection $z_1, z_2, \cdots, z_n$ with $n \ge 3$, repeatedly apply the claim so that we are left with three numbers $w_1, w_2, w_3$ inside the disk. One of these, say $w_3$, is of the form $a_1 z_1 a_2 z_2 + \cdots + a_{n-2} z_{n-2}$. Rotating the disk does not change the modulus of the sum of points in the disk, so again WLOG we can take $w_3$ to be a non-negative real.

We now need to show that we can find $a_1, a_2 = \pm 1$ so that $|w_3 + a_1 w_1 + a_2 w_2|^2 \lt 3$. Let $w_k = x_k + i y_k$. Expand out $|w_3 + a_1 w_1 + a_2 w_2|^2$ to get

$$ |w_3 + a_1 w_1 + a_2 w_2|^2 = \Big\{ x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \Big\} + 2 f(a_1, a_2) $$

where $f(a_1, a_2) = a_1 a_2 (x_1x_2 + y_1 y_2) + a_1 x_1 x_3 + a_2 x_2 x_3$. It is easy to show $f(a_1, a_2) \le 0$ for the right choices of $\pm1$ for the $a_i$.

In this case then $$ |w_3 + a_1 w_1 + a_2 w_2|^2 = \Big\{ x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \Big\} + 2 f(a_1, a_2) \le x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \le 3 $$

Answer 3

Sorry, I don't have enough reputation to comment. This is only a comment. What if you used contradiction. Say $\left| \sum a_i z_i \right| \geq \sqrt{3}$ for every choice of coefficients $a_i$. Then consider the smallest such sum (which exists since there are only finitely many possibilities). That is the end of my good idea, but it seems like you should be able to produce a smaller one, using that the sum has the form $a + bi$ with $a$ or $b > 1$. My guess is that $\sqrt{3}$ can be replaced by any number greater than $\sqrt{2}$.

Update: I like this question! Here is a different idea (but still not a proof, sorry!) Assume the claim were not true. Then let $z_1, \ldots, z_n$ denote a counter-example with minimal possible $n$. First show that $n > 2$. Then show that if $n > 2$, there is some value $\pm z_i \pm z_j$ which lies in the unit circle. We then have a contradiction to minimality, by replacing the two complex numbers $z_i, z_j$ with the single complex number $\pm z_i \pm z_j$.

Actual answer(?): We prove the claim by contradiction. Assume there exist some $z_1, \ldots, z_n$ such that every combination $\left| \sum a_i z_i \right| \geq \sqrt{3}$, and choose $z_1, \ldots, z_n$ with this property that uses the minimal possible amount of complex numbers.

First of all, we have to use at least 3 complex numbers. To see this, assume we have two complex numbers $z_1, z_2$ lying inside the unit circle. We care only about the absolute value of their sum, so we can rotate so that the bigger of the two lies on the positive real line. Then we can rescale so that the bigger of the two is exactly 1. Then we want to know what is $$\sup_{|a+bi| \leq 1} \min(|1 + a + bi|, |1-a - bi|) = \sup_{|a + bi| \leq 1} \min \sqrt{ (1 \pm a)^2 + (\pm b)^2}.$$ It is clear that this supremum is achieved when $a = 0$ and $b = \pm 1$. This corresponds to $1 \pm i$, which indeed has norm $\leq \sqrt{3}$.

Next we claim that if $z_1, z_2, z_3$ are any three complex numbers lying inside the unit circle, there exist two of them, $z_i, z_j$ and signs (not necessarily the same) such that $\left| \pm z_i \pm z_j \right| \leq 1.$ After possibly rearranging the numbers and negating some of them, we may assume $|z_1| \geq |z_2|$ and that the angle separating them is at most $\frac{\pi}{3}$ radians. We may then write $z_2 = cz_1$, where $|c| \leq 1$, and the argument of $c$ is at most $\frac{\pi}{3}$. Then $$z_1 - z_2 = (1 - c)z_1.$$ The largest possible absolute value of $1-c$ occurs when the argument of $c$ is exactly $\frac{\pi}{3}$. (I think this is clear by drawing the picture. If someone asks, I will try to write it up carefully.) Then $1 - c = (1 - \frac{1}{\sqrt{2}}) + \frac{\sqrt{3}}{2} i$. This has norm strictly less than $1$, hence the norm of $(1-c)z_1$ is also strictly less than 1.

Now return to our minimal counter-example $z_1, \ldots, z_n$. We are assuming that every sum $\sum a_i z_i$ has absolute value at least $\sqrt{3}$, and that there is no set of $n-1$ complex numbers with this property. But in the notation of the previous paragraph, $z_1 - z_2, z_3, \ldots, z_n$ must also have this property, which is a contradiction.