Question:
If the four complex numbers $z, \bar{z}, \bar{z}-2 \operatorname{Re}(\bar{z})$ and $z-2 \operatorname{Re}(z)$ represent the vertices of a square of side
4 units in the Argand plane, then $|z|$ is equal to :
Correct Option: , 4
Solution:
Let $z=x+i y$
Length of side $=4$
$\mathrm{AB}=4$
$|z-\overline{2}|=4$
$|2 y|=4 ;|y|=2$
$B C=4$
$\mid \bar{z}-(\bar{z}-2 \operatorname{Re}(\bar{z}) \mid=4$
$|2 x|=4 ;|x|=2$
$|z|=\sqrt{x^{2}+y^{2}}=\sqrt{4+4}=2 \sqrt{2}$
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