SOLUTION: (i+root3)^100 +(i-root3)^100 + 2^100=?

Question 17819: (i+root3)^100 +(i-root3)^100 + 2^100=?
Answer by kapilsinghi(68) (Show Source): You can put this solution on YOUR website!
(i+root3)^100 +(i-root3)^100 + 2^100=?
i^100 = -1^50 = 1
so (i+root3)^100 = 1(i+root3)^100
= (i^100)(i+root3)^100
= (i(i+root3))^100
= (-1 + iroot3)^100

similarly
(i-root3)^100 = (-1 - iroot3)^100

w is cuberoot of unity
w + w^2 + 1 = 0
w^3n = 1
w = (-1 - iroot3)/2
=> 2w = (-1 - iroot3)

w^2 = (-1 + iroot3)/2
2w^2 = (-1 + iroot3)
so (i+root3)^100 + (i-root3)^100 + 2^100
= (-1 + iroot3)^100 + (-1 + iroot3)^100 + 2^100
= (-1 + iroot3)^100 + (-1 + iroot3)^100 + 2^100
= (2w)^100 + (2w^2)^100 + 2^100
= (2^100)*(w^100) + (2^100)*(w^200) + 2^100
= (2^100)(w^100 + (w^200) + 1)
= (2^100)(w*w^99 + (w^2)*(w^198) + 1)
= (2^100)(w + w^2 + 1)
= (2^100)(0)
= 0

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