Question

If $\frac{{(1+i)}^2}{2-i}=x+iy,$ then $x+y$ is equal to

• A. $-\frac{2}{5}$
• B. $\frac{6}{5}$
• C. $\frac{2}{5}$
• D. $-\frac{6}{5}$

Answer 1

Answer: (C)

Given that, $\frac{{(1+i)}^2}{2-i}=x+iy$
$\Rightarrow$ $\frac{1+i^2+2i}{2-i}=x+iy$
$\Rightarrow$ $\frac{1-1+2i}{2-i}=x+iy$
$\Rightarrow$ $\frac{2i}{2-i}\times \frac{2+i}{2+i}=x+iy$
$\Rightarrow$ $\frac{2i(2+i)}{{(2)}^2+{(i)}^2}=x+iy$
$\Rightarrow$ $\frac{4i-2}{5}=x+iy$
$\Rightarrow$ $x+iy=-\frac{2}{5}+\frac{4}{5}i$
Now, $x+y=-\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$