If a > 0 and z = frac{{{{left( {1 + i} right)}^2}}}{{a - i}} , has magnitude √ {frac

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4-1.Complex numbers

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If $a > 0$ and $z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a - i}}$, has magnitude $\sqrt {\frac{2}{5}} $, then $\bar z$ is equal to:

A

$ - \frac{3}{5} - \frac{1}{5}i$

B

$ - \frac{1}{5} - \frac{3}{5}i$

C

$ - \frac{1}{5} + \frac{3}{5}i$

D

$ \frac{1}{5} - \frac{3}{5}i$

(JEE MAIN-2019)

Solution

$z=\frac{(1+i)^{2}}{a-i}=\frac{2 i(a+i)}{a^{2}+1}$

$|z|=\frac{2}{\sqrt{a^{2}+1}}=\sqrt{\frac{2}{5}} \Rightarrow a=3$

$\therefore \bar{z}=\frac{-2 i(3-i)}{10}$

$\Rightarrow \frac{-1-3 i}{5}$

Standard 11

Mathematics

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