JEE Main · 2006 · 30 Apr 2006 · Offline · Complex Numbers

Split the sum into its real and imaginary parts: it equals ${\sum }_{k=1}^{10}\sin \frac{2k\pi }{11}+i{\sum }_{k=1}^{10}\cos \frac{2k\pi }{11}$. The numbers $e^{2k\pi i/11}$ for $k=0,1,\dots ,10$ are the eleven eleventh roots of unity and add to zero, so both ${\sum }_{k=0}^{10}\sin \frac{2k\pi }{11}=0$ and ${\sum }_{k=0}^{10}\cos \frac{2k\pi }{11}=0$. Dropping the $k=0$ term, which contributes $\sin 0=0$ and $\cos 0=1$, leaves ${\sum }_{k=1}^{10}\sin \frac{2k\pi }{11}=0$ and ${\sum }_{k=1}^{10}\cos \frac{2k\pi }{11}=-1$. The value is therefore $0+i(-1)=-i$.