Question

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

Answer 1

$a+ib=\frac{{(x+i)}^2}{{2x}^2+1}$

$=\frac{x^2+i^2+2xi}{2x^2+1}$

$=\frac{x^2-1+2xi}{2x^2+1}$

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

On comparing real and imaginary parts, we obtain

$a=\frac{x^2-1}{2x^2+1}andb=\frac{2x}{2x^2+1}$

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

$=\frac{x^4+1-{2x}^2+4x^2}{{(2x^2+1)}^2}$

$=\frac{x^4+1+{2x}^2}{{(2x^2+1)}^2}$

$=\frac{{(x^2+1)}^2}{{(2x^2+1)}^2}$

$\therefore a^2+b^2=\frac{{(x^2+1)}^2}{{(2x^2+1)}^2}$
Hence, proved.