Solve the following


If $\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y$, find the value of $x^{2}+y^{2}$


$\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y \quad \ldots(1)$

$\Rightarrow\left[\frac{\left(a^{2}+1\right)^{2}}{2 a-i}\right]=\overline{x+i y}$

$\Rightarrow \frac{\left(a^{2}+1\right)^{2}}{2 a+i}=x-i y \quad \ldots(2)$

On multiplying $(1)$ and $(2)$, we get

$\frac{\left(a^{2}+1\right)^{2}}{2 a-i} \times \frac{\left(a^{2}+1\right)^{2}}{2 a+i}=(x+i y)(x-i y)$

$\Rightarrow \frac{\left(a^{2}+1\right)^{4}}{(2 a)^{2}-i^{2}}=x^{2}-i^{2} y^{2}$

$\Rightarrow \frac{\left(a^{2}+1\right)^{4}}{(2 a)^{2}+1}=x^{2}+y^{2}$

Hence, $x^{2}+y^{2}=\frac{\left(a^{2}+1\right)^{4}}{4 a^{2}+1}$

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