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

We have, $z=\frac{(1+i{)}^2}{a-i}$

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

Since  $\left|z\right|=\sqrt{2/5}$ [Given]

$\therefore \sqrt{\frac{4}{{\left(a^2+1\right)}^2}+\frac{4a^2}{{\left(a^2+1\right)}^2}}=\sqrt{\frac{2}{5}}\Rightarrow \frac{2}{\sqrt{1+a^2}}=\sqrt{\frac{2}{5}}$
$\Rightarrow \frac{4}{1+a^2}=\frac{2}{5}$                                                           [Squaring both sides]

$\begin{array}{l}\Rightarrow 1+a^2=10\Rightarrow a^2=9\Rightarrow a=\pm 3\Rightarrow a=3[\because a>0]\\ \therefore z=-\frac{1}{5}+\frac{3}{5}i\Rightarrow \overline{z}=-\frac{1}{5}-\frac{3}{5}i\end{array}$