    Question

# If z1&z2 are two complex numbers and arg(z1+z2z1−z2)=π2 but |z1+z2|≠|z1−z2|, then the figure formed by the points represented by 0, z1,z2,z1+z2 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 squarearg(z1+z2z1−z2)=π2|z1+z2|≠|z1−z2|∴ Figure formed by 0,z1z2,z1+z2⇒arg(z1+z2)−arg(z1−z2)=π2∴arg(z1+z2)=π2+arg(z1−z2)⇒z1=r1eiQ1 z2=r2eiQ2 z3=r3eiQ3 z4=r4eiQ4⇒z1+z2=r1eiQ1+r2eiQ2⇒z1−z2=r1eiQ1−r2eiQ2∴ The points represent a rhombus. Hence, the answer is a rhombus but not a square.  Suggest Corrections  0      Similar questions  Related Videos   Area under the Curve
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