If

Solution:

$\left(\frac{1+i}{1-i}\right)=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$

$=\frac{(1+i)^{2}}{1^{2}-i^{2}}$

$=\frac{1^{2}+i^{2}+2 i}{1+1} \quad\left[\because i^{2}=-1\right]$

$=\frac{1-1+2 i}{2}$

$=\frac{2 i}{2}$

$=i \quad \ldots(1)$

Also,

$\left(\frac{1-i}{1+i}\right)=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$=\frac{(1-i)^{2}}{1^{2}-i^{2}}$

$=\frac{1^{2}+i^{2}-2 i}{1+1} \quad\left[\because i^{2}=-1\right]$

$=\frac{1-1-2 i}{2}$

$=\frac{-2 i}{2}$

$=-i \quad \ldots(2)$

It is given that,

$\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$

$\Rightarrow(i)^{3}-(-i)^{3}=x+i y \quad[$ From $(1)$ and $(2)]$

$\Rightarrow i^{3}+i^{3}=x+i y$

 

$\Rightarrow 2 i^{3}=x+i y$

$\Rightarrow 0-2 i=x+i y$   $\because i^{3}=-i$

 

$\Rightarrow x=0$ and $y=-2$

Thus, $(x, y)=(0,-2)$.