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

\(z=\frac{(1+i)^2}{a+i} \)
\( =\frac{2 \mathrm{i}}{\mathrm{a}+\mathrm{i}} \)
\( =\frac{2 i(a-i)}{(a+i)(a-i)} \)
\( =\frac{2+2 \mathrm{ai}}{\mathrm{a}^2+1} \)
\( |z|=\frac{2}{\sqrt{5}} \Rightarrow \frac{4}{\left(a^2+1\right)^2}+\frac{4 a^2}{\left(a^2+1\right)^2}=\frac{4}{5} \)
\( \Rightarrow 20+20 a^2=4\left(a^4+2 a^2+1\right) \)
\( \Rightarrow 4 a^4-12 a^2-16=0 \)
\( \Rightarrow a^4-3 a^2-4=0 \)
\( \Rightarrow\left(\mathrm{a}^2-4\right)\left(\mathrm{a}^2+1\right)=0 \)
\( \Rightarrow a^2=4 \text { and } a^2=-1 \)
\( \Rightarrow \mathrm{a}=2\quad\)\(\ldots[\because a>0] \)
\( \therefore \text { (i) } \Rightarrow z=\frac{2}{5}+\frac{4}{5} \mathrm{i} \)
\( \therefore \bar{z}=\frac{2}{5}-\frac{4}{5} \mathrm{i}\)