If the four complex numbers z,overline{z},ove | vedclass

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(D) Let $z = x + iy$. Then $\overline{z} = x - iy$.$\operatorname{Re}(z) = x$ and $\operatorname{Re}(\overline{z}) = x$.The four vertices are $A(x + i

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$2 \sqrt{2}$

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(D) Let $z = x + iy$. Then $\overline{z} = x - iy$.$\operatorname{Re}(z) = x$ and $\operatorname{Re}(\overline{z}) = x$.The four vertices are $A(x + iy)$,$B(x - iy)$,$C((x - iy) - 2x) = C(-x - iy)$,and $D((x + iy) - 2x) = D(-x + iy)$.The side length of the square is given as $4$ units.The distance between $A$ and $B$ is $|(x + iy) - (x - iy)| = |2iy| = 2|y| = 4$,which implies $|y| = 2$.The distance between $B$ and $C$ is $|(x - iy) - (-x - iy)| = |2x| = 2|x| = 4$,which implies $|x| = 2$.Therefore,$|z| = \sqrt{x^2 + y^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

If the four complex numbers $z$,$\overline{z}$,$\overline{z}-2 \operatorname{Re}(\overline{z})$ 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:

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