If ω is an imaginary cube root of unity then find: (1−ω+ω2)(1−ω2+ω4)(1−ω4+ω8) ..... to 2n factors =
A
2n+1
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B
22n
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C
22n+1
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D
2n
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Solution
The correct option is B22n (1−w+w2)(1−w2+w4)(1−w4+w8)...2nterms=(1−w+w2)(1−w2+w)(1−w+w2)...=(0−2w)(0−2w2)(0−2w)(0−2w2)...2nterms=(−2w)n(−2w2)n=(2w.2w2)n=4n=22n Hence, (b) is correct.