Question

If $x+iy=\frac{(a^2+1{)}^2}{2a-i}$, what is the value of $x^2+y^2$?

Answer 1

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

To find:- $x^2+y^2=?$

Sol:-

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

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

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

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

Comparing both sides, we have

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

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

Now,

$x^2+y^2$

$={\left(2ay\right)}^2+y^2$

$=y^2(1+4a^2)$

$={\left(\frac{{(a^2+1)}^2}{4a^2+1}\right)}^2(1+4a^2)$

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