If frac{{z - α }}{{z + α }}left( {α ∈ R} right) is a purely imaginary number

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4-1.Complex numbers

hard

If $\frac{{z - \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and $\left| z \right| = 2$, then a value of $\alpha $ is

A

$2$

B

$1$

C

$\frac{1}{2}$

D

$\sqrt 2$

(JEE MAIN-2019)

Solution

$\frac{{z\, – \,\alpha }}{{z\, + \,\alpha }}\, = \, – \,\left( {\frac{{\bar z\, – \,\alpha }}{{\bar z\, + \,\alpha }}} \right)$

$z\bar z\, + \,z\alpha \, – \,\alpha \bar z\, – \,{\alpha ^2}$ $ = \, – \,z\bar z\, + \,\,\alpha \bar z\, – \,\alpha \bar z\, + {\alpha ^2}\,$

${\alpha ^2}\, = \,|z{|^2}$

$\alpha \, = \, \pm \,2$

Standard 11

Mathematics

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