If frac{{z - alpha }}{{z + alpha }}left( | vedclass

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Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\frac{{z\, - \,\alpha }}{{z\, + \,\alpha }}\, = \, - \,\left( {\frac{{\bar z\, - \,\alpha }}{{\bar z\, + \,\alpha }}} \right)$

$z\bar z\, + \,z\alpha \, - \,\alpha \bar z\, - \,{\alpha ^2}$ $ = \, - \,z\bar z\, + \,\,\alpha \bar z\, - \,\alpha \bar z\, + {\alpha ^2}\,$

${\alpha ^2}\, = \,|z{|^2}$

$\alpha \, = \, \pm \,2$

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

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