A) \[{{\tan }^{-1}}\left( \frac{2k}{{{k}^{2}}+1} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{2k}{1-{{k}^{2}}} \right)\]
C) - \[2{{\tan }^{-1}}k\]
D) \[2{{\tan }^{-1}}k\]
Correct Answer: C
Solution :
\[\left| \frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}} \right|=1\]Þ\[\frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}}=\cos \alpha +i\sin \alpha \] Þ \[\frac{2{{z}_{1}}}{-2{{z}_{2}}}=\frac{\cos \alpha +i\sin \alpha +1}{\cos \alpha -1+i\sin \alpha }\](Applying componendo and dividendo) Þ \[\frac{{{z}_{1}}}{{{z}_{2}}}=i\cot \frac{\alpha }{2}\]Þ\[i{{z}_{1}}=-\left( \cot \frac{\alpha }{2} \right)\,\,{{z}_{2}}\] But \[i{{z}_{1}}=k{{z}_{2}}\]Þ \[k=-\cot \frac{\alpha }{2}\] Now \[k=-\cot \frac{\alpha }{2}\Rightarrow \]\[\cot \frac{\alpha }{2}=-k\]Þ\[\tan \alpha =\frac{+2k}{{{k}^{2}}-1}\] Þ \[\tan \alpha =\frac{-2k}{1-{{k}^{2}}}\]Þ \[\alpha ={{\tan }^{-1}}\left( \frac{-2k}{1-{{k}^{2}}} \right)=-2{{\tan }^{-1}}k\] Now \[\frac{{{z}_{1}}-{{z}_{2}}}{{{z}_{1}}+{{z}_{2}}}=\cos \alpha +i\sin \alpha \] Þ \[\alpha \]is the angle between \[{{z}_{1}}-{{z}_{2}}\]and \[{{z}_{1}}+{{z}_{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
