Question

If $\begin{array}{rcl}{\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^3-{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^3& =& x+iy\end{array}$, then

• A. $x=0,y=-2$
• B. $x=-2,y=0$
• C. $x=1,y=1$
• D. $x=-1,y=1$

Answer 1

The correct option is A

$x=0,y=-2$

Explanation for the correct option:

STEP 1: Simplify $\begin{array}{rcl}& & {\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^{\mathbf{3}}\mathbf{-}{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^{\mathbf{3}}\end{array}$.

given $\begin{array}{rcl}{\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^3-{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^3& =& x+iy\end{array}$

$\begin{array}{rcl}{\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^3-{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^3& =& {\left[\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right]}^3-{\left[\frac{\left(1-i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}\right]}^3\\ & =& {\left[\frac{\left(1-1+2i\right)}{1+1}\right]}^3-{\left[\frac{\left(1-1-2i\right)}{1+1}\right]}^3\\ & =& {\left[\frac{2i}{2}\right]}^3-{\left[\frac{-2i}{2}\right]}^3\\ & =& -i^3-i^3\\ & =& -2i\end{array}$

STEP 3: Equate value of $\begin{array}{rcl}& & {\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^{\mathbf{3}}\mathbf{-}{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^{\mathbf{3}}\end{array}$ obtained in step 1 and step 2 to find values of $\mathit{x}\mathbf{,}\mathbf{}\mathit{y}$.

$\begin{array}{rcl}{\left[\frac{\left(1+i\right)}{\left(1-i\right)}\right]}^3-{\left[\frac{\left(1-i\right)}{\left(1+i\right)}\right]}^3& =& x+iy=0-2i\end{array}$

Comparing real and imaginary parts,

$$\begin{gathered} x=0 \\ y=-2 \end{gathered}$$

Hence, option (A) is correct.