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