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)

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

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

$\Rightarrow \frac{2\left[\,{}^{6}C_{1}i+\,{}^{6}C_{3}i^{3}+\,{}^{6}C_{5}i^{5}\right]}{2^{3}}=x+iy$

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