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Answered & Verified by Expert
$$
\text { If } \frac{(1+i)^2}{2-i}=x+i y \text {, then find the value of } x+y \text {. }
$$
\text { If } \frac{(1+i)^2}{2-i}=x+i y \text {, then find the value of } x+y \text {. }
$$
Solution:
2008 Upvotes
Verified Answer
We have, $x+i y=\frac{(1+i)^2}{2-i}$
$$
=\frac{\left(1+i^2+2 i\right)}{2-i}=\frac{2 i}{2-i}=\frac{2 i(2+i)}{(2-i)(2+i)}=\frac{4 i-2}{4+1}
$$
$$
\Rightarrow x+i y=\frac{-2}{5}+\frac{4 i}{5} \quad\left[\because i^2=-1\right]
$$
On comparing both sides, we get $x=-2 / 5$ and $y=4 / 5$
$$
\Rightarrow x+y=\frac{-2}{5}+\frac{4}{5}=2 / 5
$$
$$
=\frac{\left(1+i^2+2 i\right)}{2-i}=\frac{2 i}{2-i}=\frac{2 i(2+i)}{(2-i)(2+i)}=\frac{4 i-2}{4+1}
$$
$$
\Rightarrow x+i y=\frac{-2}{5}+\frac{4 i}{5} \quad\left[\because i^2=-1\right]
$$
On comparing both sides, we get $x=-2 / 5$ and $y=4 / 5$
$$
\Rightarrow x+y=\frac{-2}{5}+\frac{4}{5}=2 / 5
$$
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