Evaluate $\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p$ Evaluate$$\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p$$
I wanted to convert this problem in the form $e^{i\theta}$ but not able to proceed
 A: Noting $z=\cos\frac{2\pi}{11}+i\cdot \sin\frac{2\pi}{11}$ then 
$$\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}=(-i)\left(\cos\frac{2q\pi}{11}+i\sin\frac{2q\pi}{11}\right)=(-i)z^q$$
and 
$$\sum\limits_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)=(-i)\sum\limits_{q=1}^{10}z^q=(-i)\left(-1+\sum\limits_{q=0}^{10}z^q\right)=\\
(-i)\left(-1 + \frac{z^{11}-1}{z-1}\right)=(-i)\left(-1+\frac{\cos\frac{2\cdot 11\pi}{11}+i\sin\frac{2\cdot 11\pi}{11}-1}{z-1}\right)=\\
(-i)\left(-1+\frac{1-1}{z-1}\right)=i$$
Then
$$\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p=
\sum_{p=1}^{32}(3p+2)i^p=\\
3i\sum_{p=1}^{32}pi^{p-1}+2i\sum_{p=1}^{32}i^{p-1}=3i\left.\left(\sum_{p=1}^{32}px^{p-1}\right)\right|_{x=i}+2i\sum_{p=0}^{31}i^{p}=\\
3i\left.\left(\sum_{p=0}^{32}x^{p}\right)'\right|_{x=i}+2i\frac{i^{32}-1}{i-1}=
3i\left.\left(\frac{x^{33}-1}{x-1}\right)'\right|_{x=i}=
3i\left.\left(\frac{1 - 33 x^{32} + 32 x^{33}}{(x-1)^2}\right)\right|_{x=i}=
3i\left(-16-16i\right)=48(1-i)$$
A: Hint: You may consider the sum $S = \sum_{q=0}^{10}(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11})$ as tracing out a regular unit 11-gon in the complex plane, so that $S=0$.
