Question

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

• A. $1$
• B. $-1$
• C. $i$
• D. $-i$

Answer 1

The correct option is C $i$

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

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

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

$=-i\sum _{k=1}^{10}{e}^{i\frac{i2\pi k}{11}}=-i[\sum _{k=0}^{10}{e}^{i\frac{2\pi k}{11}}-1]$

$-i[\frac{e^{\frac{i2\pi }{11}}(1-e^{i2\pi })}{1-e^{\frac{i2\pi }{11}}}-1]$

$=-i\times -1=i$