(c) यहाँ $|{z_k}| = 1,k = 1,\,\,2,….n$

$⇒ |{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1}{{{z_k}}}$

इसलिए $|{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|$

वैकल्पिक : माना ${z_k} = \cos {\theta _k} + i\sin {\theta _k},\,\,\,k = 1,\,\,2,….n$

ताकि $|{z_k}| = \sqrt {{{\cos }^2}{\theta _k} + {{\sin }^2}{\theta _k}}  = 1$

 तब $\frac{1}{{{z_k}}} = {(\cos {\theta _k} + i\sin {\theta _k})^{ – 1}} = (\cos {\theta _k} – i\sin {\theta _k})$

अब ${z_1} + {z_2} + ….. + {z_n}$

$ = (\cos {\theta _1} + ….. + \cos {\theta _n}) – i(\sin {\theta _1} + ….. + \sin {\theta _n})$ और $\left( {\frac{1}{{{z_1}}}} \right) + \left( {\frac{1}{{{z_2}}}} \right) + ….. + \left( {\frac{1}{{{z_n}}}} \right)$

$ = (\cos {\theta _1} + ….. + \cos {\theta _n}) – i(\sin {\theta _1} + ….. + \sin {\theta _n})$

अत: $|{z_1} + {z_2} + ….. + {z_n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ….. + \frac{1}{{{z_n}}}} \right|$ 

प्रत्येक पक्ष $\sqrt {{{(\cos {\theta _1} + ….. + \cos {\theta _n})}^2} + {{(\sin {\theta _1} + …. + \sin {\theta _n})}^2}} $ के बराबर है।