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Question

Find the value of

6k=1 (sin2kπ7icos2kπ7).


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Solution

6k=1 (sin2kπ7icos2kπ7)

Since above expression is not a standard polar form. First convert this expression for standard polar from by taking i common

= -i 6k=1 (cos2kπ7+isin2kπ7)

We know,

Sum of n roots of Unity = 0

6k=0 (cos2kπ7+isin2kπ7) = 0

Given expression -i 6k=1 (cos2kπ7+isin2kπ7) can be written as -i[ Sum of all the roots - first root]

-i[6k=0 (cos2kπ7+isin2kπ7) - 1]

= -i[0 - 1] = i


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