Question

If the four complex numbers $z,\overline{z},\overline{z}-2Re\overline{z},z-2Re\left(z\right)$ represent the vertices of a square of side $4units$ in the Argand plane, then $\left|z\right|$is equal to:

• A. $2$
• B. $4$
• C. $4\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is D

$2\sqrt{2}$

Explanation for the correct option:

Given vertices of square are $z,\overline{z},\overline{z}-2Re\overline{z},z-2Re\left(z\right)$ and the length of side of square is $4units$

Let $z=x+iy$

$AB=4\Rightarrow \left|z-\overline{z}\right|=4$

$\Rightarrow \left|2y\right|=4\Rightarrow \left|y\right|=2$

$BC=4\Rightarrow \left|\overline{z}-\left(\overline{z}-2Re\overline{z}\right)\right|=4$

$\Rightarrow \left|2x\right|=4\Rightarrow \left|x\right|=2$

We know that $\left|z\right|=\sqrt{x^2+y^2}$

$\begin{array}{cccl}\Rightarrow & \left|z\right|& =& \sqrt{2^2+2^2}\\ & & =& \sqrt{8}\\ \Rightarrow & \left|z\right|& =& 2\sqrt{2}\end{array}$

Hence, the correct option is option(D) i.e. $2\sqrt{2}$