If $\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, find $(x, y)$
$\left(\frac{1+i}{1-i}\right)=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$
$=\frac{(1+i)^{2}}{1^{2}-i^{2}}$
$=\frac{1^{2}+i^{2}+2 i}{1+1} \quad\left[\because i^{2}=-1\right]$
$=\frac{1-1+2 i}{2}$
$=\frac{2 i}{2}$
$=i \quad \ldots(1)$
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, $(x, y)=(0,-2)$.
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