If {z_1},{z_2} are two complex numbers such t | vedclass

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Question

(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 $
==>

Vedclass · Accepted Answer

-$2{	an ^{ - 1}}k$

Vedclass · Answer

(c)$\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} ight| = 1$==>$\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)
==> $\frac{{{z_1}}}{{{z_2}}} = i\cot \frac{\alpha }{2}$==>$i{z_1} = - \left( {\cot \frac{\alpha }{2}} ight)\,\,{z_2}$
But $i{z_1} = k{z_2}$==> $k = - \cot \frac{\alpha }{2}$
Now $k = - \cot \frac{\alpha }{2} \Rightarrow $$\cot \frac{\alpha }{2} = - k$==>$	an \alpha = \frac{{ + 2k}}{{{k^2} - 1}}$
==> $	an \alpha = \frac{{ - 2k}}{{1 - {k^2}}}$==> $\alpha = {	an ^{ - 1}}\left( {\frac{{ - 2k}}{{1 - {k^2}}}} ight) = - 2{	an ^{ - 1}}k$
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 ${z_1} + {z_2}$.

If ${z_1},{z_2}$ are two complex numbers such that $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right| = 1$ and $i{z_1} = k{z_2}$, where $k \in R$, then the angle between ${z_1} - {z_2}$ and ${z_1} + {z_2}$ is

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