(c) We have $|{z_k}| = 1,k = 1,\,\,2,….n$
==> $|{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1}{{{z_k}}}$
Therefore $|{z_1} + {z_2} + …. + {z_n}| = |\overline {{z_1} + {z_2} + …. + {z_n}} |$
$(\because \,\,\,|z|\, = \,|\,\overline z |)$
$ = |{\overline z _1} + \overline {{z_2}} + ….. + {\overline z _n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + …. + \frac{1}{{{z_n}}}} \right|$
Aliter : Let ${z_k} = \cos {\theta _k} + i\sin {\theta _k},\,\,\,k = 1,\,\,2,….n$
So that $|{z_k}| = \sqrt {{{\cos }^2}{\theta _k} + {{\sin }^2}{\theta _k}} = 1$
Then $\frac{1}{{{z_k}}} = {(\cos {\theta _k} + i\sin {\theta _k})^{ – 1}} = (\cos {\theta _k} – i\sin {\theta _k})$
Now, ${z_1} + {z_2} + ….. + {z_n}$
$ = (\cos {\theta _1} + ….. + \cos {\theta _n}) – i(\sin {\theta _1} + ….. + \sin {\theta _n})$
and $\left( {\frac{1}{{{z_1}}}} \right) + \left( {\frac{1}{{{z_2}}}} \right) + ….. + \left( {\frac{1}{{{z_n}}}} \right)$
$ = (\cos {\theta _1} + ….. + \cos {\theta _n}) – i(\sin {\theta _1} + ….. + \sin {\theta _n})$
Hence $|{z_1} + {z_2} + ….. + {z_n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ….. + \frac{1}{{{z_n}}}} \right|$
Since each side is equal to
$\sqrt {{{(\cos {\theta _1} + ….. + \cos {\theta _n})}^2} + {{(\sin {\theta _1} + …. + \sin {\theta _n})}^2}} $