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: , 3
Solution:
Let $z=x+i y$
$\because$ Length of side of square $=4$ units
Then, $|z-\bar{z}|=4 \Rightarrow|2 i y|=4 \Rightarrow|y|=2$
Also, $|z-(z-2 \operatorname{Re}(z))|=4$
$\Rightarrow|2 \operatorname{Re}(z)|=4 \Rightarrow|2 x|=4 \Rightarrow|x|=2$
$\therefore|z|=\sqrt{x^{2}+y^{2}}=\sqrt{4+4}=2 \sqrt{2}$
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