Explanation

(d)
Given, $z = \cos{\theta + i\sin{\theta = e^{i\theta}}}$
$\therefore\ \ \ \sum_{m = 1}^{15}{Im\ \left( z^{2m - 1} \right) = \sum_{m = 1}^{15}{Im\ \left( e^{i\theta} \right)^{2m - 1}}}$
$= \sum_{m = 1}^{15}{Im\ e^{i(2m - 1)\theta}}$
$= \sin{\theta + \sin{3\theta + \sin{5\theta + ... + \sin{29\ \theta}}}}$
$= \frac{\sin{\left( \frac{\theta + 29\theta}{2} \right)\sin\left( \frac{15 \times 2\theta}{2} \right)}}{\sin\left( \frac{2\theta}{2} \right)}$
$= \frac{\sin{(15\ \theta)}\sin{(15\ \theta)}}{\sin\theta} = \frac{1}{4\sin{2{^\circ}}}$