Question

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

Answer 1

$$\begin{gathered} \frac{{\left(1+i\right)}^2}{2-i}=\frac{1^2+i^2+2i}{2-i} \\ =\frac{1-1+2i}{2-i}[\because i^2=-1] \\ =\frac{2i}{2-1}\times \frac{2+i}{2+i} \\ =\frac{2i(2+i)}{2^2-i^2} \\ =\frac{4i+2i^2}{4+1}[\because i^2=-1] \\ =\frac{4i-2}{5} \\ =\frac{-2}{5}+\frac{4}{5}i....\left(1\right) \end{gathered}$$


It is given that,
$$\begin{gathered} \frac{{\left(1+i\right)}^2}{2-i}=x+iy \\ \Rightarrow -\frac{2}{5}+\frac{4}{5}i=x+iy[\mathrm{From}(1\left)\right] \\ \Rightarrow x=-\frac{2}{5}\mathrm{and}y=\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \therefore x+y=\frac{-2}{5}+\frac{4}{5} \\ =\frac{2}{5} \end{gathered}$$

Thus, x + y = $\frac{2}{5}$.