Are there five complex numbers satisfying the following equalities?

Question

Can anyone help on the following question?

Are there five complex numbers $z_{1}$, $z_{2}$ , $z_{3}$ , $z_{4}$ and $z_{5}$ with $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|+\left|z_{4}\right|+\left|z_{5}\right|=1$ such that the smallest among $\left|z_{1}\right|+\left|z_{2}\right|-\left|z_{1}+z_{2}\right|$, $\left|z_{1}\right|+\left|z_{3}\right|-\left|z_{1}+z_{3}\right|$, $\left|z_{1}\right|+\left|z_{4}\right|-\left|z_{1}+z_{4}\right|$, $\left|z_{1}\right|+\left|z_{5}\right|-\left|z_{1}+z_{5}\right|$, $\left|z_{2}\right|+\left|z_{3}\right|-\left|z_{2}+z_{3}\right|$, $\left|z_{2}\right|+\left|z_{4}\right|-\left|z_{2}+z_{4}\right|$, $\left|z_{2}\right|+\left|z_{5}\right|-\left|z_{2}+z_{5}\right|$, $\left|z_{3}\right|+\left|z_{4}\right|-\left|z_{3}+z_{4}\right|$, $\left|z_{3}\right|+\left|z_{5}\right|-\left|z_{3}+z_{5}\right|$ and $\left|z_{4}\right|+\left|z_{5}\right|-\left|z_{4}+z_{5}\right|$is greater than $8/25$?

Thanks!

Answer 1

Obviously not; the sum of two of the absolute values of distinct $z_i$ is always${}\leq1$, and subtracting anything positive from it won't make the result${}>1$. In other words you won't even get one of those values you are taking the minimum of to be${}>1$.

Answer 2

Suppose you have solutions and express $z_i$ as $r_i e^{\theta_i}$.

(I use $s_i = \sin( \theta_i )$ and $c_i = \sin( \theta_i )$ to make notations shorter)

Then$$\begin{align*} |z_i| + |z_j| - |z_i + z_j| & = |r_i e^{\theta_i}| + |r_j e^{\theta_j}| - |r_i e^{\theta_i} + r_j e^{\theta_j}| \\ & = r_i + r_j - |r_i(c_i +is_i) + r_j(c_j +is_j) | \\ & = r_i + r_j - |(r_ic_i+r_jc_j) + i(r_is_i+r_js_j)| \\ & = r_i + r_j - \sqrt{(r_ic_i+r_jc_j)^2 + (r_is_i+r_js_j)^2} \\ & = r_i + r_j - \sqrt{ ( r_i^2c_i^2 + 2r_ir_jc_ic_j + r_j^2c_j^2 ) + ( r_i^2s_i^2 + 2r_ir_js_is_j + r_j^2s_j^2 ) } \\ & = r_i + r_j - \sqrt{ r_i^2( c_i^2 + s_i ^2 ) + r_j^2(c_j^2 + s_j^2) + 2r_ir_j(c_ic_j+s_is_j) } \\ |z_i| + |z_j| - |z_i + z_j| & = r_i + r_j - \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } \end{align*} $$

Then

$$\begin{align*} |z_i| + |z_j| - |z_i + z_j| > \frac{8}{25} & \Leftrightarrow r_i + r_j - \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } > \frac{8}{25} \\ & \Leftrightarrow r_i + r_j - \frac{8}{25} > \sqrt{r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) } \\ & \Leftrightarrow ( r_i + r_j - \frac{8}{25} ) ^ 2 > r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow r_i^2 + r_j^2 + (\frac{8}{25})^2 + 2r_ir_j - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > r_i^2 + r_j^2+2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow (\frac{8}{25})^2 + 2r_ir_j - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > 2r_ir_j\cos(\theta_i-\theta_j) \\ & \Leftrightarrow (\frac{8}{25})^2 + 2r_ir_j( 1 - \cos(\theta_i-\theta_j) ) - 2\frac{8}{25} r_i - 2\frac{8}{25} r_j > 0 \\ \end{align*} $$

I might update later but that's all I have for now :/

But I would try to express $r_i$ as a function of $r_{\sigma(i)}$ with $\sigma$ a permutation. And by doing that, you would probably get an ugly way of calculating any $r_j$ from all the $\theta_k$ where $j,k \in \{\sigma(i)^n, n\in \mathbb{N}\}$.

Answer 3

I think the answer is no.

Suppose we can find such $z_1,\dotsc,z_5$. Then all of them are non-zero. Let $\theta_{ij}$ be the angle between $z_i$ and $z_j$.

For all $i,j$, we have \begin{align*} \frac{8}{25} & \leq |z_i| + |z_j| - |z_i + z_j| \\ & = \frac{(|z_i| + |z_j|)^2 - |z_i + z_j|^2}{|z_i| + |z_j| + |z_i + z_j|} \\ & \leq \frac{2|z_i||z_j|(1-\cos{\theta_{ij}})}{2|z_j|} \\ & = |z_i| (1-\cos{\theta_{ij}}).\end{align*}

Reorder $z_1,\dotsc,z_5$ so that $|z_1|\leq \dotsb \leq |z_5|$. Now as $|z_1| \leq \frac{1}{5}$, the above gives $\theta_{1j} \geq \cos^{-1}\left(-\frac{3}{5}\right)=2.21$ rad. This means if we join each $z_i$ to the origin separately, there is a region of $4.42$ rad around $z_1$ without any other $z_i$'s. So $z_2,\dotsc,z_5$ must be within a region of at most $2\pi - 4.42 = 1.87$ rad. (I rounded up/down frequently so that the bounds actually become cruder at each stage.)

In particular, there is some $i \neq 1$ such that $\theta_{ij} \leq \frac{1.87}{3} = 0.63$ radians, so that $|z_i| \geq 1.67$. Contradiction.