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

Answer: (B)

$\frac{1-i}{1+i}=\frac{1-i}{1+i}\times \frac{1-i}{1-i}=\frac{1+i^2-2i}{1-i^2}=-i$
$\therefore {\left(\frac{1-i}{1+i}\right)}^{96}=a+ib$
$\Rightarrow {\left(-i\right)}^{96}=a+ib$
$\Rightarrow 1=a+ib$
$\therefore \left(a,b\right)\equiv \left(1,0\right)$