(A) We have $(1-i)^{4} = [(1-i)^{2}]^{2}$.Expanding $(1-i)^{2}$ using the identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$:$(1-i)^{2} = 1^{2} + i^{2} - 2i =

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) We have $(1-i)^{4} = [(1-i)^{2}]^{2}$.Expanding $(1-i)^{2}$ using the identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$:$(1-i)^{2} = 1^{2} + i^{2} - 2i = 1 - 1 - 2i = -2i$.Now,substitute this back into the expression:$(1-i)^{4} = (-2i)^{2}$.Calculate the square:$(-2i)^{2} = (-2)^{2} \times i^{2} = 4 \times (-1) = -4$.Thus,the complex number in the form $a+ib$ is $-4 + 0i$.