Question

The value of $\sum _{k=1}^{10}(\sin \frac{2k\pi }{11}+i\cos \frac{2k\pi }{11})$ is

• A. $1$
• B. $10i$
• C. $-i$
• D. $-10i$

Answer 1

Answer: (C)

$-i$

$s={\sum }_{k=1}^{10}\sin \frac{2k\pi }{11}+icos\frac{2k\pi }{11}=>s={\sum }_{k=1}^{10}i{e}^{\frac{-2k\pi }{11}i}=>s=i{\sum }_{k=1}^{10}{\left(e^{-\frac{2i\pi }{11}}\right)}^k=>s=i(e^{-\frac{2i\pi }{11}}+e^{-\frac{4i\pi }{11}}+....+e^{-\frac{20i\pi }{11}})=>s=i\left(\frac{e^{-\frac{2i\pi }{11}}((e^{-\frac{2i\pi }{11}}{)}^{10}-1)}{\left(e^{-\frac{2i\pi }{11}}\right)-1}\right)=>s=i\left(\frac{e^{-\frac{22i\pi }{11}}-e^{-\frac{2i\pi }{11}}}{e^{-\frac{2i\pi }{11}}-1}\right)=>s=i\left(\frac{e^{-2i\pi }-e^{-\frac{2i\pi }{11}}}{e^{-\frac{2i\pi }{11}}-1}\right)=>s=-i(\because e^{-2i\pi }=1)$