Question

The value of $\begin{array}{c}10\\ \sum \\ n=1\end{array}\left(\sin \frac{2n\pi }{11}-i\cos \frac{2n\pi }{11}\right),$ is

• A. $-1$
• B. $0$
• C. $-i$
• D. $i$

Answer 1

The correct option is B $0$

${\sum }_{n=1}^{10}(\sin \frac{2n\pi }{11}-i\cos \frac{2n\pi }{11})$

$=\frac{1}{i}{\sum }_{n=1}^{10}(isin\frac{2n\pi }{11}-i^2\cos \frac{2n\pi }{11})$

$=\frac{1}{i}{\sum }_{n=1}^{10}(\cos \frac{2n\pi }{11}+\sin \frac{2n\pi }{11})$

$=\frac{1}{i}{\sum }_{n=1}^{10}{e}^{i\frac{2n\pi }{11}}$

$=\frac{1}{i}[e^{\frac{2\pi i}{11}}+e^{\frac{4\pi }{11}i}+........e^{\frac{20\pi i}{11}}]$

Series is in $GP$ with common difference $=e^{\frac{2\pi }{11}i}$

$=\frac{1}{i}{e}^{\frac{2\pi i}{11}}\left(\frac{e^{\frac{2\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}\right)=\frac{e^{\frac{2\pi i}{11}}}{i}\left(\frac{e^{2\pi i}-1}{e^{\frac{2\pi i}{11}}-1}\right)$

$=\frac{e^{\frac{2\pi i}{11}}}{i}\left(\frac{\cos 2\pi +i\sin 2\pi -1}{e^{\frac{2\pi i}{11}}-1}\right)$

$=\frac{1}{i}(1-1)=0$