Question

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

Answer 1

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

$$=\frac{1-1+2i}{2-i}[\because i^2=-1]$$

$$=\frac{2i}{2-1}\times \frac{2+i}{2+i}$$

$$=\frac{2i(2+i)}{2^2-i^2}$$

$$=\frac{4i+2i^2}{4+1}[\because i^2=-1]$$

$$=\frac{4i-2}{5}$$

$$=\frac{-2}{5}+\frac{4}{5}i....(1)$$

It is given that,

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

$$\Rightarrow -\frac{2}{5}+\frac{4}{5}i=x+iy[\text{ From }(1)]$$

$$\Rightarrow x=-\frac{2}{5}\text{ and }y=\frac{4}{5}$$

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

$$=\frac{2}{5}$$

Thus, x + y = $$\frac{2}{5}$$.