If $z_{1}, z_{2}, z_{3}, z_{4}$ are represented by the vertices of a rhombus taken in the anticlockwise order, then

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

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z_{1} + z_{2} + z_{3} + z_{4} = 0

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a m p \frac{z_{2} - z_{4}}{z_{1} - z_{3}} = \frac{π}{2}

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a m p \frac{z_{1} - z_{2}}{z_{3} - z_{4}} = \frac{π}{2}

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Solution

The correct option is B $a m p \frac{z_{2} - z_{4}}{z_{1} - z_{3}} = \frac{π}{2}$
Since diagonals of a rhombus bisect each other
$∴ \frac{z_{1} + z_{3}}{2} = \frac{z_{2} + z_{4}}{2} = z_{0}$ (say) $\Rightarrow z_{1} - z_{2} + z_{3} - z_{4} = 0$
Also, since diagonals of a rhombus are at right angles
$∴ a m p \frac{z_{2} - z_{0}}{z_{1} - z_{0}} = \frac{π}{2}$

$\Rightarrow \frac{z_{2} - \frac{z_{2} + z_{4}}{2}}{z_{1} - \frac{z_{1} + z_{3}}{2}} = \frac{π}{2}$

$\Rightarrow a m p \frac{z_{2} - z_{4}}{z_{1} - z_{3}} = \frac{π}{2}$

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