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Maths-

General

Easy

Question

If |z₁|=|z₂| and arg⁡(z₁ )+arg⁡(z₂ )= $pi over 2$ then [ ]

z₁ z₂ is purely imaginary
z_i z₂ is purely real
$\arg (z_{1}^{- 1}) + \log (z_{2}^{- 1}) = \frac{π}{2}$
${(z_{1} + z_{2})}^{2}$ is purely real

The correct answer is: z₁ z₂ is purely imaginary

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