Question

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

Answer 1

$$\begin{gathered} \left(\frac{1+i}{1-i}\right)=\frac{1+i}{1-i}\times \frac{1+i}{1+i} \\ =\frac{{\left(1+i\right)}^2}{1^2-i^2} \\ =\frac{1^2+i^2+2i}{1+1}[\because i^2=-1] \\ =\frac{1-1+2i}{2} \\ =\frac{2i}{2} \\ =i....\left(1\right) \end{gathered}$$


Also,
$$\begin{gathered} \left(\frac{1-i}{1+i}\right)=\frac{1-i}{1+i}\times \frac{1-i}{1-i} \\ =\frac{{\left(1-i\right)}^2}{1^2-i^2} \\ =\frac{1^2+i^2-2i}{1+1}[\because i^2=-1] \\ =\frac{1-1-2i}{2} \\ =\frac{-2i}{2} \\ =-i....\left(2\right) \end{gathered}$$

It is given that,
$$\begin{gathered} {\left(\frac{1+i}{1-i}\right)}^3-{\left(\frac{1-i}{1+i}\right)}^3=x+iy \\ \Rightarrow (i{)}^3-(-i{)}^3=x+iy[\mathrm{From}(1)\mathrm{and}(2\left)\right] \\ \Rightarrow i^3+i^3=x+iy \\ \Rightarrow 2i^3=x+iy \\ \Rightarrow 0-2i=x+iy[\because i^3=-i] \\ \Rightarrow x=0\mathrm{and}y=-2 \end{gathered}$$

Thus, (x, y) = (0, −2).