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

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

Aicharya · Accepted Answer

$\begin{aligned} z & =\frac{(1+i)^2}{a-i} \ & =\frac{2 i}{a-i}=\frac{2 i(a+i)}{a^2+1} \ 	herefore \quad|z| & =\sqrt{\frac{-2+2 a i}{a^2+1}} \ \Rightarrow & \frac{2}{\sqrt{5}}=\frac{2 a^2}{\sqrt{a^2+1}}\end{aligned}$ $\Rightarrow \mathrm{a}=2 \quad \ldots[\because \mathrm{a}>0]$ $\begin{aligned} 	herefore \quad z & =\frac{-2+4 i}{5}=\frac{-2}{5}+\frac{4}{5} i \ & \Rightarrow \bar{z}=-\frac{2}{5}-\frac{4}{5} \mathrm{i}\end{aligned}$

If $\mathrm{a}>0$ and $\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}$, has…

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