Question

If ${\left(\frac{1+i}{1-i}\right)}^3-{\left(\frac{1-i}{1+i}\right)}^3=x+iy,\text{find}(x,y).$

Answer 1

$\frac{1+i}{1-i}$ $=\frac{1+i}{1-i}\times \frac{1+i}{1+i}$
$=\frac{1+i^2+2i}{1^2-i^2}$
$=i$
$\therefore \frac{1+i}{1-i}=i....\left(1\right)$

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

$\Rightarrow \frac{1-i}{1+i}=\frac{1}{i}....\left(2\right)$
$\because {\left(\frac{1+i}{1-i}\right)}^3-{\left(\frac{1-i}{1+i}\right)}^3=x+iy$
$\Rightarrow i^3-{\frac{1}{i}}^3$
$=x+iy$ {from eqn (1)& (2)}

$\Rightarrow -2i=x+iy$
$\Rightarrow x=0,y=-2$
{equating the real and imaginary parts}
$x=0,y=-2$