If $\mathrm{a}\gt0$ and $\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}$, has magnitude $\sqrt{\frac{2}{5}}$, then $\overline{\mathrm{z}}$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} z & =\frac{(1+i)^2}{a-i} \ & =\frac{2 i}{a-i} \ & =\frac{2 i(a+i)}{a^2+1} \ & =\frac{-2+2 a i}{a^2+1} \ 	herefore \quad|z| & =\sqrt{\frac{4+4 a^2}{\left(a^2+1ight)^2}} \ & \Rightarrow \sqrt{\frac{2}{5}}=\frac{2}{\sqrt{a^2+1}} \ & \Rightarrow a=3\end{aligned}$ $\begin{aligned} 	herefore \quad & z=\frac{-2+2(3) i}{3^2+1}=\frac{-2+6 i}{10} \ & \Rightarrow z=\frac{-1+3 i}{5} \ & \Rightarrow \bar{z}=\frac{-1}{5}-\frac{3 i}{5}\end{aligned}$

If $\mathrm{a}\gt0$ and $\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}$,…

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