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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(sin2kπ7−icos2kπ7)=−i∑6k=1(cos2kπ7+isin2kπ7) =−i{∑6k=1ei2kπ7}=−i{ei2kπ7+ei4kπ7+ei6kπ7+ei8kπ7+ei10kπ7+ei12kπ7} =−i⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩ei2kπ7(1−ei12π7)1−ei2π7⎫⎪
⎪
⎪⎬⎪
⎪
⎪⎭ =−i{ei2π7−ei14π71−ei2π7}[∵ei14π7=1]=−i{ei2π7−11−ei2π7}=i