Find the value of
∑6k=1 (sin2kπ7−icos2kπ7).
∑6k=1 (sin2kπ7−icos2kπ7)
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