Question

If the four complex numbers $z,\,\,\overline z ,\,\overline z - 2\operatorname{Re} \overline z ,\,z - 2\operatorname{Re} (z)$ represent the vertices of a square of side 4 units in the Argand plane, then $|z|$ is equal to
A) $2$
B) $4$
C) $4\sqrt 2 $
D) $2\sqrt 2 $

Answer 1

Hint: Draw a diagram with the four vertices as $z,\,\,\overline z ,\,\overline z - 2\operatorname{Re} \overline z ,\,z - 2\operatorname{Re} (z)$ in a way such that $z\,{\text{and}}\,\,\overline z - 2\operatorname{Re} \overline z $; $\overline z \,{\text{and}}\,z - 2\operatorname{Re} (z)$ are the two pairs of diagonally opposite points. After finding x and y from the diagram, use the formula \[|z|\, = \,\sqrt {{x^2} + {y^2}} \].

Complete step by step Solution:
Let $z = x + iy$,
Then, \[\overline z = x - iy\,\], $\overline z - 2\operatorname{Re} \overline z = x - iy - 2x = - x - iy$ and $z - 2\operatorname{Re} (z) = x + iy - 2x = - x + iy$
Representing this as a diagram we get,

It is given to us that each side is 4 units. Therefore,
$AB = BC = CD = DA = 4\,{\text{units}}$
$AB = \,|z - \overline z |\, = \,|2y|\, = 4$ units
$|y|\, = \,2$ units
$BC = \,|\overline z - (\overline z - 2\operatorname{Re} (\overline z ))|\, = \,|2x|\, = 4$ units
$|x|\, = \,2$ units
$|z|\, = \,\sqrt {{x^2} + {y^2}} = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 $ units

Hence, the correct option is (D).

Note: Care must be taken while choosing the diagonally opposite points. In the diagram, $A$ cannot be diagonally opposite to $B$ or $D$. We can also find the values of $x$ and $y$ from $CD$ and $DA$ instead of $AB$ and $BC$.