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Question

The value of 6k=1(sin2kπ7i cos2kπ7) is

A
-1
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B
\N
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C
-i
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D
i
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Solution

The correct option is D i
6k=1(sin2kπ7i cos2kπ7)=i6k=1(cos2kπ7+i sin2kπ7)
=i{6k=1ei2kπ7}=i{ei2kπ7+ei4kπ7+ei6kπ7+ei8kπ7+ei10kπ7+ei12kπ7}
=i⎪ ⎪ ⎪⎪ ⎪ ⎪ei2kπ7(1ei12π7)1ei2π7⎪ ⎪ ⎪⎪ ⎪ ⎪
=i{ei2π7ei14π71ei2π7}[ei14π7=1]=i{ei2π711ei2π7}=i

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