Question

If $(1+i{)}^3/(1-i{)}^3-(1-i{)}^3/(1+i{)}^3=x+iy,$ then

• A. $x=0$, $y=-2$
• B. $x=-2$, $y=0$
• C. $x=1$, $y=1$
• D. $x=-1$, $y=1$

Answer 1

Answer: (A)

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

$\Rightarrow \frac{{\left(1+i\right)}^6-{\left(1-i\right)}^6}{{\left[1-{\left(i\right)}^2\right]}^3}=x+iy$

$\Rightarrow \frac{2\left[{}^6{C}_{1}i+{}^6{C}_3{i}^3+{}^6{C}_5{i}^5\right]}{2^3}=x+iy$

$\Rightarrow \frac{1}{2^2}\left[6i-20i+6i\right]=x+iy$
$\Rightarrow -2i+x+iy$
$\Rightarrow x=0,y=-2$.