Question

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

Answer 1

Given that : $\frac{(a^2+1{)}^2}{2a-1}=x+iy$

$\Rightarrow \frac{(a^2+1{)}^2}{(2a-i)}=x+iy\Rightarrow \frac{(a^2+1{)}^2(2a+i)}{(2a-i)(2a+i)}=x+iy\Rightarrow \frac{(a^2+1{)}^2(2a+i)}{4a^2+1}=x+iy\Rightarrow \frac{2a(a^2+1{)}^2}{4a^2+1}\text{and}y=\frac{(a^2+1{)}^2}{4a^2+1}\therefore x+y^2=4a^2{\left[\frac{(a^2+1{)}^2}{4a^2+1}\right]}^2+{\left[\frac{(a^2+1{)}^2}{4a^2+1}\right]}^2=\frac{(4a^2+1)(a^2+1{)}^4}{(4a^2+1{)}^2}=\frac{(a^2+1{)}^4}{(4a^2+1)}$