Explanation

(c)
Given, $\left( \cos{\theta + i\sin\theta} \right)\left( \cos{2\theta + i\sin{2\theta}} \right)\ldots$
$\left( \cos{n\theta + i\sin{n\theta}} \right) = 1$
$\therefore\ \ \cos{(\theta + 2\theta + 3\theta + ... + n\theta)} + i\sin{(\theta + 2\theta + 3\theta + ... + n\theta) = 1}$
$\Rightarrow \ \ \cos{\left( \frac{n(n + 1)}{2}\theta \right) + i\sin{\left( \frac{n(n + 1)}{2}\theta \right) = 1}}$
$\Rightarrow \ \ \ \ \cos{\left( \frac{n(n + 1)}{2}\theta \right) = 1}$ and $\sin\left( \frac{n(n + 1)}{2}\theta \right) = 0$
$\therefore\ \ \ \frac{n(n + 1)}{2}\theta = 2m\pi\ \Rightarrow \ \ \theta = \frac{4m\pi}{n(n + 1)}$