Question

The value of $\displaystyle \sum_{k=1}^6 \bigg(\sin\frac{2\pi k}{7}-i \cos \frac{2\pi k}{7}\bigg)$ is

• A. -1
• B. 0
• C. -i
• D. i

Answer 1

Answer: (D)

$\displaystyle \sum_{k=1}^6\bigg(\sin\frac{2\pi k}{7}-i \cos \frac{2\pi k}{7}\bigg)=\displaystyle \sum_{k=1}^6 -i \bigg(\cos\frac{2\pi k}{7}+i \sin \frac{2\pi k}{7}\bigg)$
$=-i\bigg \{ \displaystyle \sum_{k=1}^6 e^{\frac{i2k\pi}{7}}\bigg\}=i \{ e^{i2\pi/7}+e^{i4\pi/7}+e^{i6\pi/7}$
$ \, +e^{i8\pi/7}+e^{i10\pi/7}+e^{i12\pi/7} \}$
$=-i \bigg \{e^{i2\pi/7}\frac{(1-e^{i12\pi/7})}{1-e^{i2\pi/7}}\bigg \}$
$=-i \bigg \{ \frac{e^{i2\pi/7}-e^{i14\pi/7}}{1-e^{i2\pi/7}}\bigg\} [\because \, e^{i14\pi/7=1}]$
$=-i\bigg \{ \frac{e^{i2\pi/7-1}}{1-e^{i 2\pi/7}}\bigg\}=i$