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Addition of Vectors

If z1 & z2 ...

Question

If $z_{1} & z_{2}$ are two complex numbers and $a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}$ but $| z_{1} + z_{2} | \neq | z_{1} - z_{2} |$ , then the figure formed by the points represented by 0, $z_{1}, z_{2}, z_{1} + z_{2}$ is

A

a parallelogram but not a rectangle or a rhombus

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B

a rectangle but not a square

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C

a rhombus but not a square

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D

a square

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Solution

The correct option is C a rhombus but not a square
$a r g (\frac{z_{1} + z_{2}}{z_{1} - z_{2}}) = \frac{π}{2}$
$| z_{1} + z_{2} | \neq | z_{1} - z_{2} |$
$∴$ Figure formed by $0, z_{1} z_{2}, z_{1} + z_{2}$
$\Rightarrow a r g (z_{1} + z_{2}) - a r g (z_{1} - z_{2}) = \frac{π}{2}$
$∴ a r g (z_{1} + z_{2}) = \frac{π}{2} + a r g (z_{1} - z_{2})$
$\Rightarrow z_{1} = r_{1} e^{i Q_{1}}$ $z_{2} = r_{2} e^{i Q_{2}}$ $z_{3} = r_{3} e^{i Q_{3}}$ $z_{4} = r_{4} e^{i Q_{4}}$
$\Rightarrow z_{1} + z_{2} = r_{1} e^{i Q_{1}} + r_{2} e^{i Q_{2}}$
$\Rightarrow z_{1} - z_{2} = r_{1} e^{i Q_{1}} - r_{2} e^{i Q_{2}}$
$∴$ The points represent a rhombus.
Hence, the answer is a rhombus but not a square.

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