(c) $\left| {\frac{{{z_1} – {z_2}}}{{{z_1} + {z_2}}}} \right| = 1$

$⇒\frac{{{z_1} – {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha  + i\sin \alpha $

$\frac{{2{z_1}}}{{ – 2{z_2}}} = \frac{{\cos \alpha  + i\sin \alpha  + 1}}{{\cos \alpha  – 1 + i\sin \alpha }}$

(योगान्तरानुपात से)

$⇒\frac{{{z_1}}}{{{z_2}}} = i\cot \frac{\alpha }{2}$

$⇒i{z_1} =  – \left( {\cot \frac{\alpha }{2}} \right)\,\,{z_2}$

लेकिन $i{z_1} = k{z_2}⇒ k =  – \cot \frac{\alpha }{2}$

अब $k =  – \cot \frac{\alpha }{2} \Rightarrow $$\cot \frac{\alpha }{2} =  – k$Þ$\tan \alpha  = \frac{{ + 2k}}{{{k^2} – 1}}$

$⇒  \tan \alpha  = \frac{{ – 2k}}{{1 – {k^2}}}$

$⇒ \alpha  = {\tan ^{ – 1}}\left( {\frac{{ – 2k}}{{1 – {k^2}}}} \right) =  – 2{\tan ^{ – 1}}k$

अब $\frac{{{z_1} – {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha  + i\sin \alpha $

$⇒{z_1} – {z_2}$ एवं ${z_1} + {z_2}$के बीच का कोण $\alpha $ है