Let $4 \sum i = 1 a_{i} = 0$ and $4 \sum j = 1 a_{j} z_{j} = 0$ and $a_{1} a_{2} | z_{1} - z_{2} |^{2} = a_{3} a_{4} | z_{3} - z_{4} |^{2}$ .
Then show that $z_{1}, z_{2}, z_{3}, z_{4}$ are concyclic

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Solution

$∵ 4 \sum i = 1 a_{i} = 0$
$\Rightarrow a_{1} + a_{2} + a_{3} + a_{4} = 0$
$\Rightarrow (a_{1} + a_{3}) = - (a_{2} + a_{4})$ ........ (1)
and $4 \sum j = 1 a_{j} z_{j} = 0$
$∴ a_{1} z_{1} + a_{2} z_{2} + a_{3} z_{3} + a_{4} z_{4} = 0$
$\Rightarrow (a_{1} z_{1} + a_{3} z_{3}) = - (a_{2} z_{2} + a_{4} z_{4})$ ..... (2)
Dividing (2) by (1) we get
$\frac{a_{1} z_{1} + a_{3} z_{3}}{a_{1} + a_{3}} = \frac{a_{2} z_{2} + a_{4} z_{4}}{a_{2} + a_{4}}$ ...... (3)
(3) Represents point O divides PR in the ratio $a_{3} : a_{1}$ and O divides QS in the ratio $a_{2} : a_{4}$
Let $O R = a_{1} k, O P = a_{3} k, O Q = a_{4} l, O S = a_{2} l$
Now In $Δ$ OPQ
$(P Q)^{2} = (O P)^{2} + (O Q)^{2} - 2 (O P) (O Q) c o s θ$
$\Rightarrow | z_{1} - z_{2} |^{2} = a_{3}^{2} k^{2} + a_{4}^{2} l^{2} - 2 a_{3} a_{4} l k c o s θ$
$∴ a_{1} a_{2} | z_{1} - z_{2} |^{2} = a_{1} a_{2} a_{3}^{2} k^{2} + a_{1} a_{2} a_{4}^{2} l^{2} - 2 a_{1} a_{2} a_{3} a_{4} l k c o s θ$
Similarly, $a_{3} a_{4} | z_{3} - z_{4} |^{2} = a_{3} a_{4} a_{1}^{2} k^{2} + a_{3} a_{4} a_{2}^{2} l^{2} - 2 a_{1} a_{2} a_{3} a_{4} l k c o s θ$
from given condition
$a_{1} a_{2} | z_{1} - z_{2} |^{2} = a_{3} a_{4} | z_{3} - z_{4} |^{2}$
$∴ a_{1} a_{2} a_{3}^{2} k^{2} a_{1} a_{2} a_{4}^{2} l^{2} = a_{3} a_{4} a_{1}^{2} k^{2} + a_{3} a_{4} a_{2}^{2} l^{2}$
$\Rightarrow k^{2} a_{3} a_{1} (a_{2} a_{3} - a_{1} a_{4}) = l^{2} a_{2} a_{4} (a_{2} a_{3} - a_{1} a_{4})$
$\Rightarrow (a_{1} k) (a_{3} k) = (a_{2} l) (a_{4} l)$
$O P . O R = O Q . O S$
So P, Q, R and S are concyclic.

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| z_{1} | = | z_{2} | = | z_{3} | = | z_{4} | = 1

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z_{1} + z_{2} + z_{3} + z_{4} = 0

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z_{1} + z_{2} + z_{3} + z_{4} = 0

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| z_{1} - z_{2} |^{2} + | z_{2} - z_{3} |^{2} + | z_{3} - z_{4} |^{2} + | z_{4} - z_{1} |^{2} - 2 (| z_{1} |^{2} + | z_{2} |^{2} + | z_{3} |^{2} + | z_{4} |^{2})

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If $z_{1}, z_{2}$ are conjugate complex numbers, and $z_{3}, z_{4}$ are also conjugate, then show that

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If | $z_{1} | = | z_{2} | = | z_{3} | = | z_{4} | = 1$ and $z_{1} + z_{2} + z_{3} + z_{4} = 0$
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Let 4∑i=1ai=0 and 4∑j=1ajzj=0 and a1a2|z1−z2|2=a3a4|z3−z4|2. Then show that z1,z2,z3,z4 are concyclic

Let $4 \sum i = 1 a_{i} = 0$ and $4 \sum j = 1 a_{j} z_{j} = 0$ and $a_{1} a_{2} | z_{1} - z_{2} |^{2} = a_{3} a_{4} | z_{3} - z_{4} |^{2}$ .
Then show that $z_{1}, z_{2}, z_{3}, z_{4}$ are concyclic