Question

$\dfrac{z - \alpha }{z + \alpha }$
is purely imaginary and |Z| =B2 then
$| \alpha | \: is$
​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

it has given that, (z - α)/(z + α) is purely imaginary and |z| = 82

To find : the value of |α|

solution : concept : if z is purely imaginary,

then $z+\bar{z}=0$

so, $\frac{z-\alpha}{z+\alpha}+\frac{\bar{z-\alpha}}{\bar{z+\alpha}}=0$

⇒$\frac{z-\alpha}{z+\alpha}+\frac{\bar{z}-\bar\alpha}{\bar{z}+\bar\alpha}=0$

⇒$(z-\alpha)(\bar z+\bar\alpha)+(\bar z-\bar\alpha)(z+\alpha)=0$

⇒$z\bar z+z\bar\alpha-\alpha\bar z-\alpha\bar\alpha+\bar z z+\bar z\alpha-\bar\alpha z-\bar\alpha\alpha=0$

we know, $z\bar z=|z|^2$ and $\alpha\bar\alpha=|\alpha|^2$

⇒2|z|² - 2|α|² = 0

⇒|z| = |α|

but |z| = 82

so, |α| = 82

Therefore the value of |α| = 82