Question

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

Answer 1

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

$L.H.S.=\frac{(1+i{)}^2}{2-i}$
$\Rightarrow L.H.S.=\frac{1+i^2+2i}{2-i}$
$\Rightarrow L.H.S.=\frac{1-1+2i}{2-i}\{\because i^2=-1\}$
$\to L.H.S.=\frac{2i}{2-i}$

Multiplying and dividing by
conjugate of denominator in $L.H.S.$
$\Rightarrow L.H.S.=\frac{2i(2+i)}{(2-i)(2+i)}=\frac{4i+2i^2}{4-i^2}=\frac{4i-2}{4-(-1)}$
$\Rightarrow L.H.S.=\frac{-2+4i}{5}$
$\Rightarrow L.H.S.=\frac{-2}{5}+\frac{4}{5i}$

Comparing $L.H.S.$ and $R.H.S.$
$\because L.H.S.=R.H.S.$
$\therefore \frac{-2}{5}+\frac{4}{5i}=x+iy$

Comparing corresponding real and imaginary part
$\Rightarrow x=\frac{-2}{5},y=\frac{4}{5}$

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