Prove that

Solution:

Consider

$x+i y=\left(\frac{1+i}{1-i}\right)^{93}-\left(\frac{1-i}{1+i}\right)^{3}$

Now, rationalizing

$x+i y=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{93}-\left(\frac{1-i}{1+i} \times \frac{1-i}{1-i}\right)^{3}$

$=\left(\frac{(1+i)^{2}}{(1-i)(1+i)}\right)^{93}-\left(\frac{(1-i)^{2}}{(1+i)(1-i)}\right)^{3}$

In denominator, we use the identity

$(a-b)(a+b)=a^{2}-b^{2}$

$=\left(\frac{1+i^{2}+2 i}{(1)^{2}-(i)^{2}}\right)^{93}-\left(\frac{1+i^{2}-2 i}{(1)^{2}-(i)^{2}}\right)^{3}$

$=\left(\frac{1+(-1)+2 i}{1-i^{2}}\right)^{93}-\left(\frac{1+(-1)-2 i}{1-i^{2}}\right)^{3}$

$=\left(\frac{2 i}{1-(-1)}\right)^{93}-\left(\frac{-2 i}{1-(-1)}\right)^{3}$

$=\left(\frac{2 i}{2}\right)^{93}-\left(\frac{-2 i}{2}\right)^{3}$

$=(\mathrm{i})^{93}-(-\mathrm{i})^{3}$

$=(\mathrm{i})^{92+1}-\left[-(\mathrm{i})^{3}\right]$

$=\left[(\mathrm{i})^{92}(\mathrm{i})\right]-\left[-\left(\mathrm{i}^{2} \times \mathrm{i}\right)\right]$

$=\left[\left(\mathrm{i}^{4}\right)^{23}(\mathrm{i})\right]-[-(-\mathrm{i})]$

$=\left[(1)^{23}(\mathrm{i})\right]-\mathrm{i}$

$=\mathrm{i}-\mathrm{i}$

x + iy = 0

$\therefore x=0$ and $y=0$