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1,w,w^2.....w^n-1 are nth roots of unity then (1-w)(1-w^2)...(1-w^n-1) =?

1,w,w^2.....w^n-1 are nth roots of unity then (1-w)(1-w^2)...(1-w^n-1) =?

Grade:12

3 Answers

Arun
25750 Points
4 years ago
x=1 on both the sides nowAns =3x^2 + x +1= (x-a_1)(x-a_2).....(x-a_{n-1})\frac{x^3 -1}{(x-1)}= (x-a_1)(x-a_2).....(x-a_{n-1})x^3 -1= (x-1)(x-a_1)(x-a_2).....(x-a_{n-1}) – 1 =0So I can write3Basically 1,a1,a2......an-1 all these are the roots of this equation x
 
Aditya Gupta
2081 Points
4 years ago
1,w,w^2.....w^n-1 are nth roots of unity means these are roots of polynomial eqn:
x^n – 1= 0
or x^n – 1= (x – 1)(x – w)(x – w^2)..........(x – w^n-1)
or (x – w)(x – w^2)..........(x – w^n-1)= (x^n – 1)/(x – 1)
now, take limit x tends to 1 both sides
we get (1-w)(1-w^2)...(1-w^n-1) = lim x tends to 1 (x^n – 1)/(x – 1)
apply l hospital rule (0/0)
= lim x tends to 1 nx^n-1/1
= n
kindly approve :))
Vikas TU
14149 Points
4 years ago

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