Explanation

(c)
We have, $\mathcal{z}_{r} = \cos{\frac{r\alpha}{n^{2}} + i\sin\frac{r\alpha}{n^{2}}}$
where $r = 1,\ 2,\ 3,\ldots,n$
$\therefore\ \ \mathcal{z}_{1} = \cos{\frac{\alpha}{n^{2}} + i\sin{\frac{\alpha}{n^{2}};}}$
$\mathcal{z}_{2} = \cos{\frac{2\alpha}{n^{2}} + i\sin\frac{2\alpha}{n^{2}}}$
$\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
$\mathcal{z}_{n} = \cos{\frac{n\alpha}{n^{2}} + i\sin\frac{\alpha}{n^{2}}\ }$
$\therefore\ \ \lim_{n \rightarrow \infty}{(\mathcal{z}_{1}\mathcal{z}_{2}\mathcal{z}_{3}\ldots\mathcal{z}_{n})}$
$= \lim_{n \rightarrow \infty}\left( \cos{\frac{\alpha}{n^{2}} + i\sin\frac{\alpha}{n^{2}}} \right)\left( \cos{\frac{2\alpha}{n^{2}} + i\sin\frac{2\alpha}{n^{2}}} \right)\ldots\left( \cos{\frac{n\alpha}{n^{2}} + i\sin\frac{n\alpha}{n^{2}}} \right)$
$= \lim_{n \rightarrow \infty}\left\lbrack \cos\left\{ \frac{\alpha}{n^{2}}(1 + 2 + 3 + ... + n) \right\} \right.\ + \ i\sin\left. \ \left\{ \frac{\alpha}{n^{2}}(1 + 2 + 3 + ... + n) \right\} \right\rbrack$
$= \lim_{n \rightarrow \infty}\left\lbrack \cos{\frac{\alpha n(n + 1)}{2n^{2}} + i\sin\frac{\alpha n(n + 1)}{2n^{2}}} \right\rbrack$
$= \cos{\frac{\alpha}{2} + i\sin\frac{\alpha}{2}}$
$= e^{\frac{i\alpha}{2}}$