(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 $
==> $\frac{{2{z_1}}}{{ – 2{z_2}}} = \frac{{\cos \alpha + i\sin \alpha + 1}}{{\cos \alpha – 1 + i\sin \alpha }}$
(Applying componendo and dividendo)
==> $\frac{{{z_1}}}{{{z_2}}} = i\cot \frac{\alpha }{2}$==>$i{z_1} = – \left( {\cot \frac{\alpha }{2}} \right)\,\,{z_2}$
But $i{z_1} = k{z_2}$==> $k = – \cot \frac{\alpha }{2}$
Now $k = – \cot \frac{\alpha }{2} \Rightarrow $$\cot \frac{\alpha }{2} = – k$==>$\tan \alpha = \frac{{ + 2k}}{{{k^2} – 1}}$
==> $\tan \alpha = \frac{{ – 2k}}{{1 – {k^2}}}$==> $\alpha = {\tan ^{ – 1}}\left( {\frac{{ – 2k}}{{1 – {k^2}}}} \right) = – 2{\tan ^{ – 1}}k$
Now $\frac{{{z_1} – {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha + i\sin \alpha $
==> $\alpha $ is the angle between ${z_1} – {z_2}$ and ${z_1} + {z_2}$.