If ω is an imaginary cube root of unity, then the value of (1-ω+ω2)⋅(1-ω2+ω4)⋅(1-ω4+ω8) dots (2n factors ) is

Question

If ω is an imaginary cube root of unity, then the value of (1-ω+ω2)⋅(1-ω2+ω4)⋅(1-ω4+ω8) dots (2n factors ) is (A) 22n (B) 2n (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

Given, ω3=1,1+ω+ω2=0...(i) Now, (1-ω+ω2) ⋅(1-ω2+ω4) ⋅(1-ω4+ω8) (1-ω8+ω16)...2 n factors =(1-ω+ω2)(1-ω2+ω) ⋅(1-ω+ω2) ⋅(1-ω2+ω)...2 n factors [from Eq.(i)] =(-ω-ω) ⋅(-ω2-ω2) ⋅(-ω-ω) ⋅(-ω2-ω2)...2 n factors [from Eq. (i)] =(-2 ω) ⋅(-2 ω2) ⋅(-2 ω) ⋅(-2 ω2)... 2 n factors =(-2)2 n(ω)n(ω2)n =(-1)2 n(2)2 n ω3 n =22 n ⋅(ω3)n=22 n(1)n=22 n [from Eq .(i)]

Q. If $ω$ is an imaginary cube root of unity, then the value of $(1 - ω + ω^{2}) \cdot (1 - ω^{2} + ω^{4}) \cdot (1 - ω^{4} + ω^{8}) \dots$ (2n factors ) is

$2^{2 n}$

$2^{n}$

$1$

$0$

Q. If ω is an imaginary cube root of unity, then the value of (1−ω+ω2)⋅(1−ω2+ω4)⋅(1−ω4+ω8)… (2n factors ) is

22n

2n

1

0

Q. If $ω$ is an imaginary cube root of unity, then the value of $(1 - ω + ω^{2}) \cdot (1 - ω^{2} + ω^{4}) \cdot (1 - ω^{4} + ω^{8}) \dots$ (2n factors ) is

$2^{2 n}$

$2^{n}$

$1$

$0$