Question

If $z_1,z_2,z_3,z_4$ are the vertices of a square in the argand plane, then prove that $2z_2=(1-i)z_1+(1+i)z_3$

Answer 1

Given:$z_1,z_2,z_3,z_4$ are the vertices of a square in the argand plane.

We need $z_2$ in the form of $z_1$ and $z_3$

So we rotate about $B.$

Rotation about $B$ in clockwise direction:

$z_1-z_2=(z_3-z_2)e^{\frac{i\pi }{2}}$

$\Rightarrow z_1-z_2=(z_3-z_2)(\cos \frac{\pi }{2}-i\sin \frac{\pi }{2})$

$\Rightarrow z_1-z_2=(z_3-z_2)(-i)$

$\Rightarrow z_1-z_2=-i{z}_3+i{z}_2$

$\Rightarrow z_1+i{z}_3=+i{z}_2+z_2$

$\Rightarrow \frac{z_1}{1+i}+\frac{i{z}_3}{i+1}=z_2$

$\Rightarrow z_2=\frac{z_1}{1+i}\times \frac{1-i}{1-i}+\frac{i{z}_3}{i+1}\times \frac{1-i}{1-i}$

$\Rightarrow \frac{z_1(1-i)}{2}+\frac{z_3(1+i)}{2}=z_2$

$\Rightarrow 2z_2=z_1(1-i)+(1+i)z_3$

Hence proved.