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Question

The value of S=6k=1(sin2πk7icos2πk7) ?

A
1
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B
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C
i
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D
i
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Solution

The correct option is D i
sin2πk7icos2πk7
=i(cos2πk7+isin2πk7)
=ie2πk/7i
S=6k=1i(e2πk/i)
=i(6k=1ei2πk/7)
=i [sum of roots of the equation
x6+x5+x4+x3+x2+1=0].
=i(1)
=i

1177083_1039505_ans_0d7a47a0f59e4258b0c3e756c7ebe7e8.jpg

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