Question:
if $\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}=x+i y$, , then find $(x, y)$.
Solution:
According to the question,
We have,
$x+i y=\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}$
$=\left(\frac{(1+i)^{2}}{1-i^{2}}\right)^{3}-\left(\frac{(1-i)^{2}}{1-i^{2}}\right)^{3}$
$=\left(\frac{1+2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{3}-\left(\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{3}$
$=\left(\frac{2 \mathrm{i}}{2}\right)^{3}-\left(\frac{-2 \mathrm{i}}{2}\right)^{3}$
$=i^{3}-\left(-i^{3}\right)$
$=2 i^{3}$
$=0-2 i$
Thus, $(x, y)=(0,-2)$
= i3 – (-i3)
= 2i3
= 0 – 2i
Thus, (x, y) = (0, -2)