If $a+ib = \frac{(x+i)^{2}}{2x^{2}+1}$,prove that $a^{2}+b^{2} = \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$.

(N/A) Given $a+ib = \frac{(x+i)^{2}}{2x^{2}+1}$.
Expanding the numerator using $(A+B)^{2} = A^{2}+B^{2}+2AB$:
$a+ib = \frac{x^{2}+i^{2}+2xi}{2x^{2}+1}$
Since $i^{2} = -1$:
$a+ib = \frac{x^{2}-1+i(2x)}{2x^{2}+1}$
Separating the real and imaginary parts:
$a+ib = \frac{x^{2}-1}{2x^{2}+1} + i\left(\frac{2x}{2x^{2}+1}\right)$
Comparing real and imaginary parts:
$a = \frac{x^{2}-1}{2x^{2}+1}$ and $b = \frac{2x}{2x^{2}+1}$
Now,calculate $a^{2}+b^{2}$:
$a^{2}+b^{2} = \left(\frac{x^{2}-1}{2x^{2}+1}\right)^{2} + \left(\frac{2x}{2x^{2}+1}\right)^{2}$
$= \frac{(x^{2}-1)^{2} + (2x)^{2}}{(2x^{2}+1)^{2}}$
$= \frac{x^{4}+1-2x^{2}+4x^{2}}{(2x^{2}+1)^{2}}$
$= \frac{x^{4}+2x^{2}+1}{(2x^{2}+1)^{2}}$
$= \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
Thus,$a^{2}+b^{2} = \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$. Hence,proved.