If $z_{1}, z_{2}, z_{3}, z_{4}$ are the vertices of a square in that order, then which of the following is wrong?

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$\frac{z_{1} - z_{3}}{z_{2} - z_{4}}$ is real

$|z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{4}| = |z_{4} - z_{1}|$

$z_{1} + z_{3} = z_{2} + z_{4}$

$|z_{1} - z_{3}| = |z_{2} - z_{4}|$

answer is D.

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Detailed Solution

Let the four points represented by $z_{1}, z_{2}, z_{3}, z_{4}$ in the Agrand plane be A,B,C,D respectively.

Since ABCD is a square, midpoint of AC = midpoint of BD.

$i . e, \frac{1}{2} (z_{1} + z_{3}) = \frac{1}{2} (z_{2} + z_{4}) o r z_{1} + z_{3} = z_{2} + z_{4 .} S i n c e A B = B C = C D = D A w e h a v e |z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{4}| = |z_{4} - z_{1}| S i n c e t h e d i a g o n a l s o f t h e s q u a r e a r e e q u a l A B = C D, w e h a v e |z_{1} - z_{3}| = |z_{2} - z_{4}| S i n c e A C ⊥ B D, (z_{1} - z_{3}) (z_{2} - z_{4}) i s p u r e i m a g i n a r y$