Question

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

Answer 1

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

Rationalizing, we get,

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

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

$(i{)}^3-(-i{)}^3=x+iy$

$-i-(+i)=x+iy$

$-2i=x+iy$

so, $y=-2$

$x=0$