Question

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

• A. 2i
• B. -2i
• C. 2
• D. 1

Answer 1

Answer: (A)

Let $E=-2 i \displaystyle\sum_{p=1}^{6}\left(\cos \frac{2 p \pi}{7}+i \sin \frac{2 p \pi}{7}\right)$
$E=-2 i \displaystyle\sum_{p=1}^{6} e^{i\left(\frac{2 p \pi}{7}\right)} $
Let $\alpha=e^{i\left(\frac{2 \pi}{7}\right)} \ldots( i ) $
$\therefore E=-2 i\left(\alpha+\alpha^{2}+\ldots+\alpha^{6}\right)$
$\Rightarrow E=-2 i\left[\left(1+\alpha+\alpha^{2}+\ldots+\alpha^{6}\right)-1\right] $
$E=-2 i\left[\left(\frac{1-\alpha^{7}}{1-\alpha}\right)-1\right]$
Since, $\alpha^{7}=\left(e^{i\left(\frac{2 \pi}{7}\right)}\right)^{7}=e^{i 2 \pi}=1$
$\Rightarrow E=-2 i\left\{\frac{1-\left(e^{i \frac{2 \pi}{7}}\right)^{7}}{1-e^{i\left(\frac{2 \pi}{7}\right)}}-1\right\}$
$\Rightarrow E=-2 i\left\{\frac{1-1}{1-e^{i\left(\frac{2 \pi}{7}\right)}}-1\right\}$
$\Rightarrow E=-2 i(0-1)=2 i$