Question

The value of $ \displaystyle\sum_{r=1}^{16}\left(\sin\frac{2r\pi}{17}+i\,\cos \frac{2r\pi}{17}\right) $ is

• A. $1$
• B. $i$
• C. $- i$
• D. $- 1$

Answer 1

Answer: (C)

We have $\displaystyle\sum_{r=1}^{16}$ $\left(sin \frac{2r\,\pi}{17}+i\,cos \frac{2r\,\pi}{17}\right)$
$=i \displaystyle\sum_{r=1}^{16}$ $\left(cos \frac{2r\,\pi}{17}+i\,sin \frac{2r\,\pi}{17}\right)$
$=i\left(z+z^{2}+......+z^{16}\right)$
[where $z= cos \frac{2\,\pi}{17}-i\,sin \frac{2\,\pi}{17}$]
$=i \frac{z\left(1-z^{16}\right)}{1-z}=\frac{i\left(z-z^{17}\right)}{1-z}=\frac{i\left(z-1\right)}{1-z}=i$
$\left[\because z^{17}=1\right]$