value of $a_1|z_1|^2+a_2|z_2|^2+a_3|z_3|^2+a_4|z_4|^2$

Question

If $z_{1},z_{2},z_{3},z_{4}$ are concyclic and

$a_{1}+a_{2}+a_3+a_4=0$

and $a_1z_1+a_2z_2+a_3z_3+a_4z_4=0$.

Then value of

$a_1|z_1|^2+a_2|z_2|^2+a_3|z_3|^2+a_4|z_4|^2=$,

where $a_1,a_2,a_3,a_4\in\mathbb{R}$

Here from figure, We have

$\displaystyle \frac{z_2-z_3}{z_1-z_3}=\frac{BC}{AC}e^{i\theta}$

and $\displaystyle \frac{z_2-z_4}{z_1-z_4}=\frac{BD}{AD}e^{i\theta}$

Now we have $\displaystyle (\frac{z_2-z_3}{z_1-z_3})(\frac{z_1-z_4}{z_2-z_4})=\frac{BC}{AC}\cdot \frac{AD}{BD}=\text{real}$

$\displaystyle (\frac{z_2-z_3}{z_1-z_3})\frac{(z_1-z_4}{z_2-z_4})=k\;, k\in\mathbb{R}$

And here using $|z|^2=z\bar{z}$

So $a_1|z_1|^2+a_2|z_2|^2+a_3|z_3|^2+a_4|z_4|^2$

$\displaystyle =a_1z_1\bar{z_1}+a_2z_2\bar{z_2}+a_3z_3\bar{z_3}+a_4z_4\bar{z_4}$

How do i solve it help me.

Answer 1

Let $w\in \mathbb{C}$ be the circumcenter of $z_1,\dots,z_4$, and $R$ the circumradius.

Then

$$\sum_i a_i\vert z_i-w\vert^2 =\sum a_iR^2 = 0,$$

Expanding the sum on the LHS, \begin{align*} 0=\sum_i a_i\vert z_i-w\vert^2 &= \sum_i a_i\vert z_i\vert^2 - \sum_i a_iz_i\bar w - \sum_i a_iw\bar z_i + \sum_i a_i\vert w \vert^2\\ &=\sum_i a_i\vert z_i\vert^2 - \bar w \left(\sum_i a_iz_i\right) - w\left(\sum_i a_i\bar z_i\right) + \vert w \vert^2\left({\sum_i a_i}\right)\\ &= \sum_i a_i\vert z_i \vert^2. \end{align*}