Mathematics - Complex Number Question with Solution | TestHub

We know that, $\frac{z_2-z_1}{z_4-z_1}=\frac{|z_2-z_1|}{|z_4-z_1|}{e}^{i\pi /2}=i$

(as $\left|z_2-z_1\right|=\left|z_4-z_1\right|$ )

$\Rightarrow {\left(z_4-z_1\right)}^2+{\left(z_2-z_1\right)}^2=0$

similarly $\frac{z_4-z_3}{z_2-z_3}=i$

$\Rightarrow {\left(z_4-z_3\right)}^2+{\left(z_2-z_3\right)}^2=0$

On adding Eqs (i) and (ii) , we get $2\left[z_1^2+z_2^2+z_3^2+z_4^2-z_1{z}_2-z_4{z}_1-z_4{z}_3-z_2{z}_3\right]=0$ $z_1^2+z_2^2+z_3^2+z_4^2=z_1{z}_2+z_2{z}_3+z_3{z}_4+z_4{z}_1$