Solve the following


If $\frac{\left(a^{2}+1\right)^{2}}{2 a-i}=x+i y$, then $x^{2}+y^{2}$ is equal to

(a) $\frac{\left(a^{2}+1\right)^{4}}{4 a^{2}+1}$

(b) $\frac{(a+1)^{2}}{4 a^{2}+1}$

(c) $\frac{\left(a^{2}-1\right)^{2}}{\left(4 a^{2}-1\right)^{2}}$

(d) none of these


(a) $\frac{\left(a^{2}+1\right)^{4}}{4 a^{2}+1}$

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

Taking modulus on both the sides, we get:

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

Squaring both sides, we get,

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

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