Question

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

Answer 1

$\because \frac{(a^2+1{)}^2}{(2a-i)}=x+iy$

Taking modulus of both the sides
$\Rightarrow \left|\frac{(a^2+1{)}^2}{(2a-i)}\right|=\mid x+iy\mid$

$\Rightarrow \frac{\mid (a^2+1{)}^2\mid }{\mid 2a-i\mid }=\mid x+iy\mid$

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

Squaring both sides
$\Rightarrow \frac{(a^2+1{)}^4}{4a^2+1}=x^2+y^2$

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