If a < 0 and z = (1 + i)2 / a - i,


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

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

  2. $-\frac{1}{5}+\frac{3}{5} \mathrm{i}$

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

  4. $\frac{1}{5}-\frac{3}{5} \mathrm{i}$

Correct Option: , 3


Given $a>0$

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

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

So $\bar{z}=\frac{-2 i(3-i)}{10}=\frac{-1-3 i}{5}$

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