Question

If $\begin{array}{rcl}\frac{{\left(1+i\right)}^2}{\left(2-i\right)}& =& x+iy\end{array}$, then $x+y$ is

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

Answer 1

The correct option is C

$\frac{2}{5}$

Explanation for the correct option:

STEP 1: Multiply and divide $\begin{array}{rcl}& & \frac{{(1+\mathrm{i})}^2}{(2-\mathrm{i})}\end{array}$ by $\mathrm{(}\mathrm{2}\mathrm{+}\mathrm{i}\mathrm{)}$.

Given, $\begin{array}{rcl}\frac{{\left(1+i\right)}^2}{\left(2-i\right)}& =& x+iy\end{array}$

$\begin{array}{rcl}\frac{{\left(1+i\right)}^2}{\left(2-i\right)}& =& \frac{{\left(1+i\right)}^2\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}\\ & =& \frac{\left(1-1+2i\right)\left(2+i\right)}{\left(4+1\right)}\\ & =& \frac{2i\left(2+i\right)}{5}\\ & =& \frac{4i-2}{5}\\ & =& \frac{-2}{5}+\frac{4i}{5}\end{array}$

STEP 2: Equate value of $\begin{array}{rcl}& & \frac{{\left(1+i\right)}^2}{\left(2-i\right)}\end{array}$ obtained in step 1 and step 2 to find values of $x,y$.

$\begin{array}{rcl}\frac{{\left(1+i\right)}^2}{\left(2-i\right)}& =& x+iy\\ & \Rightarrow & \frac{-2}{5}+\frac{4i}{5}=x+iy\end{array}$

Comparing real and imaginary parts of both sides, we get

$$\begin{gathered} x=\frac{-2}{5} \\ y=\frac{4}{5} \end{gathered}$$

STEP 3: Find the value of $\mathit{x}\mathbf{+}\mathit{y}$.

$\begin{array}{rcl}\therefore x+y& =& \frac{-2}{5}+\frac{4}{5}\\ & =& \frac{\mathbf{2}}{\mathbf{5}}\end{array}$

Hence option C is correct.