Question 17359
(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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