If ((1-i)/(1+i))^100 = a+ib, then (1) a = 2, b = -1 (2) a = 1, b = 0 (3) a = 0, b = 1 (4) a = -1, b = 2

Solution:

((1-i)/(1+i)) = ((1-i)(1-i)/(1+i)(1-i))

= (1-2i-1)/(1+1)

= -2i/2

= -i

((1-i)/(1+i))¹⁰⁰ = -i¹⁰⁰

= (i⁴)²⁵

= 1

Comparing with a+ib, a = 1, b = 0

Hence option (2) is the answer.