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)

(1,0)

Answer 2

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