FInd the value of ∑7k=0 (cos2kπ7+isin2kπ7)
We know,Sum of the n roots of unity = 0
∑6k=0 (cos2kπ7+isin2kπ7) = 0
∑7k=0 cos2kπ7+isin2kπ7 = ∑6k=0 (cos2kπ7+isin2kπ7) + cos2π +isin2π
= 0 + 1 = 1
Find the value of
∑6k=1 (sin2kπ7−icos2kπ7).
If the point (2k – 3, k + 2) lies on the line 2x + 3y + 15 = 0, find the value of k.