The value of ∑4k=1 (sin2π k7−icos2π k7) is
Let z=cos2π7+isin2π7, then by De Moiver's theorem Zk=cos2π k7+isin2π k7
Now the given sum
s=∑6k=1(sin2π k7−icos2π k7)
=∑6k=1[(−i)(cos2π k7+isin2π k7)]
=(−i)∑6k=1(cos2π k7+isin2π k7)=(−i)∑6k=1zk
which is a G.P. of which the first term is z, number of the terms is 6 and the common
ratio is z=cos2π7+isin2π7≠1 So summing up the G.P., we have
S=(−i)z(1−z6)1−z=(−i)z−z71−z=(−i)z−11−z=i
[∵z7=(cos2π7+isin2π7)7=cos2π+isin2π=1