Complex Numers

ok. then

\sum_{q=1}^{10}{sin(2q\pi /11)-icos(2q\pi /11)}

=-i\sum_{q=1}^{10}{cos(2q\pi /11)+isin(2q\pi /11)}

= i

So the sum is =\sum_{p=1}^{32}{(3p+2)i^p}=\sum_{p=1}^{32}{(3p)i^p}

= 3i(1-3+5-7+9-11+...-31) + 3(-2+4-6+8-10+12-...+32)
= 3i(-16) + 3(16)
= 48(1-i)

you made a silly mistake...........the real part should be +ve........the correct answer is 48(1-i).
Check your the IIIrd last step again.....

yes.. was in a hurry edited