Proving $\sum_{k=1}^{4} b_k |z_k|^2=0$ if $z_k$ are con-cyclic s. t. $\sum_{k=1}^4 b_k =0=\sum_{k=1}^{4} b_k z_k$

Question

If four complex numbers $z_k$ are con-cyclic in Argand plane such that $\sum_{k=1}^4 b_k =0=\sum_{k=1}^{4} b_k z_k$ and $b_k$ are real, earlier it has been proved that $$b_1b_2|z_1-z_2|^2=b_3 b_4|z_3-z_4|^2~~~~~(1)$$ after a good number of steps. See Condition for cyclic quadrilateral given $\sum_{i=1}^{4}{b_i}=0$ and $\sum_{i=1}^{4}{b_iz_i}=0$

In a rather simple way one can prove that $$\sum_{k=1}^{4} b_k |z_k|^2=0,~~~~(2)$$ Since $z_k$ are con-cyclic $|z_k-z_0|^2=R^2$, where let $z_0$ denote the center of the circle. Then we can write $$0=\sum_{k=1}^{4} b_k=\sum_{k=1}^{4} b_k R^2=\sum_{k=1}^4 b_k|z_k-z_0|^2 \implies \sum_{k=1}^{4} b_k [|z_k|^2+|z_0|^2-2 \Re (\bar z_0 z_k)]=0$$ $$ \implies \sum_{k=1}^{4} b_k |z_k|^2-\sum_{k=1}^42 b_k \Re(\bar z_0 z_k)=0$$ as $\sum_{k=1}^{4} b_k=0$ $$\implies \sum_{k=1}^4 b_k |z_k|^2-2\Re \left(\bar z_0 \sum_{k=1}^{4} b_k z_k\right)=0 \implies \sum_{k=1}^4 b_k |z_k|^2=0.$$

The question is how else the result (2) can be proved using (1) and its proof or otherwise.

Answer 1

The following will prove that $(1) \Leftrightarrow (2)$ under the given assumptions.

To simplify the calculations, first note that $(2)$ is invariant to translations $z \to z-w$ since $\sum b_k|z_k - w|^2=\sum b_k|z_k|^2$ following an argument similar to the one used in the original post. The assumptions are obviously invariant to translations, and so is $(1)$.

Also, both assumptions and $(1),(2)$ are invariant to complex multiplication $z \to z \cdot r\,e^{i \varphi}$.

It can then be assumed WLOG that $z_4 = 0$ and $z_3=1$ by combining a translation that makes $z_4$ the origin, and a rotation with appropriate scaling that brings $z_3$ to the unit point on the real axis.

With that, the assumptions reduce to:

$$ b_1 + b_2 + b_3 = -b_4 \tag{a} $$ $$b_1 z_1 + b_2 z_2 = -b_3 \tag{b} $$

And the relations $(1),(2)$ reduce to:

$$ b_1b_2|z_1-z_2|^2 = -b_3(b_1+b_2+b_3) \tag{1} $$ $$ b_1|z_1|^2+b_2|z_2|^2 + b_3 = 0 \tag{2} $$

Multiplying assumption $(b)$ by its conjugate gives:

$$ \begin{align} b_3^2 &= \left(b_1z_1+b_2z_2\right)\left(b_1\bar z_1+b_2\bar z_2\right) \\ &= b_1^2|z_1|^2 + b_2^2|z_2|^2 + 2 b_1b_2 \Re{(z_1\bar z_2)} \end{align} $$ $$ \implies 2 b_1b_2 \Re{(z_1\bar z_2)} = b_3^2 - b_1^2|z_1|^2 - b_2^2|z_2|^2 $$

Then:

$$ \begin{align} b_1b_2|z_1-z_2|^2 &= b_1b_2\left(|z_1|^2+|z_2|^2-2\Re{(z_1\bar z_2)}\right) \\ &= b_1b_2|z_1|^2 + b_1b_2|z_2|^2 - \left(b_3^2 - b_1^2|z_1|^2 - b_2^2|z_2|^2\right) \\ &= \left(b_1b_2|z_1|^2+b_2^2|z_2|^2\right) + \left(b_1^2|z_1|^2+b_1b_2|z_2|^2\right) - b_3^2 \\ &= b_2\left(b_1|z_1|^2+b_2|z_2|^2\color{red}{+b_3-b_3}\right)+b_1\left(b_1|z_1|^2+b_2|z_2|^2\color{red}{+b_3-b_3}\right) - b_3^2 \\ &= (b_1+b_2)\left(b_1|z_1|^2+b_2|z_2|^2+b_3\right) - b_3\left(b_1+b_2+b_3\right) \end{align} $$

It follows that $(1) \Leftrightarrow (2)$ when $b_1+b_2 \ne 0$, and the case $b_1+b_2 = 0$ can be easily verified.