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Question

The value of 6k=1(sin(2πk/7)icos(2πk/7)) is

A
-1
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B
0
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C
-i
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D
i
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E
none
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Solution

The correct option is C i
6k=1(sin(2πk7)icos(2πk7))
=(i)6k=1(cos(2πk7)isin(2πk7))
=(i)[6k=0(cos(2πk7)isin(2πk7))cos0+isin0]
=(i)[6k=0(cos(2πk7)isin(2πk7))1]

=(i)[01]=i (Sum of nth roots of unity is zero)

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