Prove the following
Question:

If

$\frac{(1+i)^{2}}{2-i}=x+i y$, , then find the value of $x+y$

Solution:

According to the question,

We have,

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

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

$=\frac{2 \mathrm{i}}{2-\mathrm{i}}$

Rationalizing the denominator,

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

$=\frac{4 \mathrm{i}+2 \mathrm{i}^{2}}{4-\mathrm{i}^{2}}$

$=\frac{4 \mathrm{i}-2}{4+1}$

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

Thus,

$\mathrm{X}=-\frac{2}{5}, \mathrm{y}=\frac{4}{5}$

Hence,

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