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Question

If ω is an imaginary cube root of unity then find: (1ω+ω2)(1ω2+ω4)(1ω4+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 B 22n
(1w+w2)(1w2+w4)(1w4+w8)...2nterms=(1w+w2)(1w2+w)(1w+w2)...=(02w)(02w2)(02w)(02w2)...2nterms=(2w)n(2w2)n=(2w.2w2)n=4n=22n
Hence, (b) is correct.

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