Question

If (1-i/1+i)^96 =a+ib then (a,b) is ​
A) (1,1)
B) (1,0)
C) (0,1)
D) (0,-1)

Answer 1

Answer:

(a , b) = (1 , 0) is the correct answer

Step-by-step explanation:

We are given

a+ib = ( (1 - i ) / (1 + i) )^96

       = ( (1 - i )(1 - i) / (1 + i)( 1 - i) )^96       ∵ By rationalization

       = ( (1 - i )² / (1² - i²) )^96                  ∵ (a² - b²) = (a + b)(a - b)

       =  ( (1² + i² - 2i ) / (1 + 1) )^96          ∵ (a - b)² = a² + b² - 2ab

       = ( (1 - 1 - 2i ) / (2) )^96                   ∵ i² = -1

       = ( ( - 2i ) / (2) )^96              

       = ( ( - i ) )^96

       = ( - i )^96

       = (-1)^96 (i)^96       ∵ (-1)^96 = -1 because 96 is even

       = i^96

       = (i²)^48

       = (-1)^48                   ∵ (-1)^96 = -1 because 96 is even

       = 1

       = 1 + 0(i)

So

a + bi = 1 + (0)i

Thus

(a , b) = (1 , 0) is the correct answer