If `((1+i)/(1-i))^3-((1-i)/(1+i))^3=x+i y ,\ fin d\ (x , y)`

Explanation

Given that , `((1+i)/(1-i))^(3)-((1-i)/(1+i))^(3) =x+iy " " ….(i) `
`:." " ((1+i)/(1-i))^(3)=(1+i^(3)+i(1+i))/(1-i^(3)-3i(1-i))=(1-i-3i+3i^(2))/(1+i-3i+3i^(2)`
`=(2i-2)/(-2i-2)=(i-1)/(-i-1)=(1-i)/(1+i)`
`((1-i))/((1+i))((1-i))/((1-i))=(1+i^(2)-2i)/(1+1)=(1-1-2i)/(2)`
`rArr" "((1+i)/(1-i))^(3)=-i " "...(ii)`
Similarly, `" "((1+i)/(1-i))^(3)=(-1)/i=(i^(2))/(i)=i " "...(iii)`
Using Eqs. (ii) and (iii) in Eq. (i), we get
`-i-i=x+i y`
`rArr -2i=x+i y`
On comparing real and imaginary part of complex of complex number, we get
`x = 0 and y = -2
So, (x,y) = 0, -2)`