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If w is an imaginary cube root of unity ,then find the value of (1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5).................(1+w^3n)

Profile image of Suraj Kumar
6 Years agoGrade
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1 Answer

Profile image of Vikas TU
6 Years ago
Dear student 
w is cube root of unity so, 
w^3 =1 
1+w+w^2 = 0 
(1+w)(1+w^2) = 1+ w+ w^2 + w^3 
w^3 = 1
(1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5).................(1+w^3n)
(1+w)(1+w^2)(1+w^3)(1+w^4)(1+w^5)(1+w^6)
=> 2*2 
We have n pairs of same brackets 
so , Ans = 2* 2* 2.......n times 
=> 2^n