Explanation

(c)
We have, $\left| \mathcal{z}_{k} \right| = 1,\ k = 1,\ 2\ldots,n$
$\Rightarrow \ \ \left| \mathcal{z}_{k} \right|^{2} = 1\ \Rightarrow \ \mathcal{z}_{k}{\overline{\mathcal{z}}}_{k} = 1\ \Rightarrow \ {\overline{\mathcal{z}}}_{k} = \frac{1}{\mathcal{z}_{k}}\$
$\therefore\ \ \left| \mathcal{z}_{1} + \mathcal{z}_{2} + ... + \mathcal{z}_{n} \right| = \left| \overline{\mathcal{z}_{1} + \mathcal{z}_{2} + ... + \mathcal{z}_{n}} \right|\ \ \ \ (\because\left| \mathcal{z} \right| = \left| \overline{\mathcal{z}} \right|)$
$= \left| {\overline{\mathcal{z}}}_{1} + {\overline{\mathcal{z}}}_{2} + ... + {\overline{\mathcal{z}}}_{n} \right|$
$= \left| \frac{1}{\mathcal{z}_{1}} + \frac{1}{\mathcal{z}_{2}} + ... + \frac{1}{\mathcal{z}_{n}} \right|$