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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

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    $\begingroup$ Transform $\sin\frac{2q\pi}{11}-i \cos\frac{2q\pi}{11}=-i(\cos\frac{2q\pi}{11}+i\sin\frac{2q\pi}{11})=-ie^{i\tfrac{2q\pi}{11}}$ $\endgroup$ – Jean Marie Nov 26 '17 at 12:30
  • $\begingroup$ No it is -i$e^\frac{2q\pi}{11}$ $\endgroup$ – Samar Imam Zaidi Nov 26 '17 at 12:34
  • $\begingroup$ @Jean Marie i used it but could not procced $\endgroup$ – Samar Imam Zaidi Nov 26 '17 at 12:35
  • $\begingroup$ You mean $-ie^{i\tfrac{2q\pi}{11}}$ it is what I have written. $\endgroup$ – Jean Marie Nov 26 '17 at 12:37
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    $\begingroup$ @Jean Marie initially did'nt see your full comment so extremaly sorry $\endgroup$ – Samar Imam Zaidi Nov 26 '17 at 12:40
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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 $$\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p= \sum_{p=1}^{32}(3p+2)i^p=\\ 3i\sum_{p=1}^{32}pi^{p-1}+2i\sum_{p=1}^{32}i^{p-1}=3i\left.\left(\sum_{p=1}^{32}px^{p-1}\right)\right|_{x=i}+2i\sum_{p=0}^{31}i^{p}=\\ 3i\left.\left(\sum_{p=0}^{32}x^{p}\right)'\right|_{x=i}+2i\frac{i^{32}-1}{i-1}= 3i\left.\left(\frac{x^{33}-1}{x-1}\right)'\right|_{x=i}= 3i\left.\left(\frac{1 - 33 x^{32} + 32 x^{33}}{(x-1)^2}\right)\right|_{x=i}= 3i\left(-16-16i\right)=48(1-i)$$

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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 $S=0$.

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  • $\begingroup$ I thought of it had it been q but 2q is a spoiler $\endgroup$ – Samar Imam Zaidi Nov 26 '17 at 12:55

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