Question

If $\frac{{\left(a^2+1\right)}^2}{2a-i}=x+iy$, then $x^2+y^2$ is equal to

• A. $\frac{{\left(a^2+1\right)}^4}{4a^2+1}$
• B. $\frac{{\left(a+1\right)}^2}{4a^2+1}$
• C. $\frac{\left(a^2-1^2\right)}{{\left(4a^2-1\right)}^2}$
• D. None of these

Answer 1

The correct option is A $\frac{{\left(a^2+1\right)}^4}{4a^2+1}$

Let, $z=x+iy$,

Then, $|z|=\sqrt{x^2+y^2}⟹|z{|}^2=x^2+y^2$

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

$⟹z=\frac{(a^2+1{)}^2}{2a-i}$

Hence, $x^2+y^2=|z{|}^2=|\frac{(a^2+1{)}^2}{2a-i}{|}^2$

$x^2+y^2=|z{|}^2=\frac{|(a^2+1{)}^2{|}^2}{|2a-i{|}^2}=\frac{(a^2+1{)}^4}{4a^2+1}$.