Question

If ${\left(\frac{1-i}{1+i}\right)}^{96}=a+ib$, then $(a,b)$ is

• A. $(1,1)$
• B. $(1,0)$
• C. $(0,1)$
• D. $(0,-1)$

Answer 1

The correct option is B $(1,0)$

$\frac{1-i}{1+i}=\frac{(1-i{)}^2}{(1+i)(1-i)}$
$=\frac{-2i}{2}=-i$
$\therefore {\left(\frac{1-i}{1+i}\right)}^{96}=(-i{)}^{96}=(i^4{)}^{24}=1=1+i\cdot 0$
$\therefore (a,b)=(1,0)$