Question

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

Answer 1

$$\begin{gathered} \frac{{\left(a^2+1\right)}^2}{2a-i}=x+iy....\left(1\right) \\ \Rightarrow \left[\overline{\frac{{\left(a^2+1\right)}^2}{2a-i}}\right]=\overline{x+iy} \\ \Rightarrow \frac{{\left(a^2+1\right)}^2}{2a+i}=x-iy....\left(2\right) \\ \mathrm{On}\mathrm{multiplying}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{{\left(a^2+1\right)}^2}{2a-i}\times \frac{{\left(a^2+1\right)}^2}{2a+i}=\left(x+iy\right)\left(x-iy\right) \\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{{\left(2a\right)}^2-i^2}=x^2-i^2{y}^2 \\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{{\left(2a\right)}^2+1}=x^2+y^2 \end{gathered}$$

Hence, $x^2+y^2=\frac{{\left(a^2+1\right)}^4}{4a^2+1}$.