The value of (i+√3)^100+(i-√3)^100+2^100 =?
Let a = (i+√3), and b = (i-√3).
We want to find the value of a^100 + b^100.
So then,
a^100 + b^100 = (a^50 + b^50)^2 – 2(ab)^50 (1)
a^50 + b^50 = (a^25 + b^25)^2 – 2(ab)^25 (2)
a^25 + b^25 = (a^5)^5 + (b^5)^5 (3)
Also,
a^5 = (i+√3)^5 = 16.(i-√3) à a^5 = 2^4.b (4)
b^5 = (i+√3)^5 = 16.(i+√3) à b^5 = 2^4.a (5)
a^5 = 2^4.b (6)
b^5 = 2^4.a (7)
And,
ab = (i+√3) (i+√3) = -1 – 3 = -4 = -2^2
ab = -2^2
Using (6) and (7) in (3),
a^25 + b^25 = (2^4.b)^5 + (2^4.a)^5
a^25 + b^25 = 2^20.(b^5 + a^5)
a^25 + b^25 = 2^20.( 16.(i+√3) + 16. (i-√3)), using (4) and (5).
a^25 + b^25 = 2^20.( 32.i) = 2^20.(2^5.i)
a^25 + b^25 = 2^25.i (8)
substituting for (8) in (2)
a^50 + b^50 = (2^25.i)^2 – 2(ab)^25
a^50 + b^50 = 2^50.i^2 – 2(ab)^25
a^50 + b^50 = -2^50 – 2(-2^2)^25
a^50 + b^50 = -2^50 + 2.2^50
a^50 + b^50 = 2^50 (9)
substituting for (9) in (1),
a^100 + b^100 = (a^50 + b^50)^2 – 2(ab)^50
a^100 + b^100 = (2^50)^2 – 2(-2^2)^50
a^100 + b^100 = 2^100 – 2(2^100)
a^100 + b^100 = -2^100
So, (i+√3)^100+(i-√3)^100+2^100 = -2^100 + 2^100 = 0