Explanation

(d)
Let $\mathcal{z =}\cos{\frac{2\pi}{7} + i\sin\frac{2\pi}{7}\ }$, then by using De Moivre’s theorem
$\therefore\ \mathcal{z}^{k} = \cos\frac{2\pi k}{7} + i\sin\frac{2\pi k}{7}$ …(i)
Let $S = \sum_{k = 1}^{6}\left( \sin{\frac{2\pi k}{7} - i\cos\frac{2\pi k}{7}} \right)$
$= \sum_{k = 1}^{6}\left\lbrack ( - i)\left( \cos{\frac{2\pi k}{7} + i\sin\frac{2\pi k}{7}} \right) \right\rbrack$
$= ( - i)\sum_{k = 1}^{6}\left( \cos{\frac{2\pi k}{7} + i\sin\frac{2\pi k}{7}} \right)$
$= ( - i)\sum_{k = 1}^{6}\mathcal{z}^{k}$ [from Eq.(i)]
$= ( - i)\mathcal{\lbrack z +}\mathcal{z}^{2} + \mathcal{z}^{3} + ... + \mathcal{z}^{6}\rbrack$
It is GP of which the first term is $\mathcal{z,}$ number of terms is 6 and the common ratio is
$\mathcal{z =}\cos{\frac{2\pi}{7} + i\sin{\frac{2\pi}{7} \neq 1}}$
$\therefore\ \ S = ( - i)\frac{\mathcal{z}(1 - \mathcal{z}^{6})}{1 - \mathcal{z}}$
$= ( - i)\frac{\mathcal{z -}\mathcal{z}^{7}}{1 - \mathcal{z}}$
$= ( - i)\frac{\mathcal{z -}\mathcal{z}^{7}}{1 - \mathcal{z}} = i\ \ \ \left\lbrack \begin{array}{r}
\because\ \mathcal{z}^{7} = \left( \cos{\frac{2\pi}{7} + i\sin\frac{2\pi}{7}} \right)^{7}\ \\
= \cos{2\pi + i\sin{2\pi = 1}}
\end{array} \right\rbrack$