If `1,alpha,alpha^2,alpha^3,......,alpha^(n-1)`are `n` `n^(th)` roots of unity, then find the value of `(2011-alpha)(2011-alpha^2)....(2011-alpha^(n-1))`

Explanation

Let `rootn(1)=x`
`Rightarrow x^(n) -1 =0, " which has n roots " , alpha, alpha^(2),......alpha^(n-1), say`
`Rightarrow x^(n-1)-= (x-1)(x-alpha) (x-alpha^(2))(x-alpha^(3)) ......(x -alpha^(n-1))=0`
`Rightarrow (x^(n)-1)/(x-1) = (x -alpha) (x-alpha^(2)) (x -alpha^(3))....(x-alpha^(n-1))`
Putting , x = 2011 , we get
` ( 2011 -alpha)(2011 -alpha^(2))(2011-alpha^(3))....(2011 -alpha^(n-1))= ((2011)^(n)-1)/2010`
we can generalise the result , as ` ( m -alpha) , ( m -alpha^(2)) ( m -alpha^(3)) ...... ( m -alpha^(n-1)) = ( m^(n) -1)/(m-1) , m ne 1`
In particular when we take lim ` x to 1`
`(1-alpha) (1 -alpha^(2))(1-alpha^(3)).....(1-alpha^(n-1))= lim_(xto1).(x^(n) -1)/(x-1) =n`
`(m + alpha) (m +alpha^(2))(m + alpha^(3)) .....(m +alpha^(n-1)) = ( 1 -(-m)^(n))/(1+m) .(-1)^(n-1)`