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Question Number 17775 by gourav~ last updated on 10/Jul/17

$$\underset{m=1}{\sum ^{10}}(\frac{\sin 2\pi m}{11}-i\frac{\cos 2\pi m}{11})=?????$$$$Solve..it$$$$$$

Answered by alex041103 last updated on 10/Jul/17

$$\underset{m=1}{\sum ^{10}}(\frac{\sin 2\pi m}{11}-i\frac{\cos 2\pi m}{11})=$$$$=\frac{1}{11}\underset{\mathrm{m}=1}{\sum ^{10}}\left(\sin 2\pi \mathrm{m}\right)-\frac{\mathrm{i}}{11}\underset{\mathrm{m}=1}{\sum ^{10}}\left(\cos 2\pi \mathrm{m}\right)$$$$\mathrm{Because}\mathrm{m}\in \mathbb{Z}$$$$\Rightarrow \sin 2\pi \mathrm{m}=0\mathrm{and}\cos 2\pi \mathrm{m}=1$$$$\underset{m=1}{\sum ^{10}}(\frac{\sin 2\pi m}{11}-i\frac{\cos 2\pi m}{11})=$$$$=\frac{1}{11}\underset{\mathrm{m}=1}{\sum ^{10}}\left(0\right)-\frac{\mathrm{i}}{11}\underset{\mathrm{m}=1}{\sum ^{10}}\left(1\right)$$$$=0\times \frac{1}{11}-\frac{\mathrm{i}}{11}\times \left(1\right)\times 10$$$$=-\frac{10}{11}\mathrm{i}$$$$\underset{\mathrm{m}=1}{\sum ^{10}}(\frac{\sin 2\pi \mathrm{m}}{11}-\mathrm{i}\frac{\cos 2\pi \mathrm{m}}{11})=-\frac{10}{11}\mathrm{i}$$$$$$$$$$

Answered by ajfour last updated on 10/Jul/17

$$=-\frac{i}{11}\underset{\mathrm{m}=0}{\sum ^{10}}{e}^{i\left(2\pi \mathrm{m}\right)}=-\frac{i}{11}\left(10\right)=-\frac{10i}{11}.$$

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