Complex Numbers And Quadratic Equations question 73
Question: If $ z_1,z_2 $ are two complex numbers such that $ | \frac{z_1-z_2}{z_1+z_2} |=1 $ and $ iz_1=kz_2 $ , where $ k\in R $ , then the angle between $ z_1-z_2 $ and $ z_1+z_2 $ is
Options:
A) $ {{\tan }^{-1}}( \frac{2k}{k^{2}+1} ) $
B) $ {{\tan }^{-1}}( \frac{2k}{1-k^{2}} ) $
C) - $ 2{{\tan }^{-1}}k $
D) $ 2{{\tan }^{-1}}k $
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Answer:
Correct Answer: C
Solution:
$ | \frac{z_1-z_2}{z_1+z_2} |=1 $
Þ $ \frac{z_1-z_2}{z_1+z_2}=\cos \alpha +i\sin \alpha $
Þ $ \frac{2z_1}{-2z_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 $ iz_1=-( \cot \frac{\alpha }{2} )z_2 $ But $ iz_1=kz_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}}( \frac{-2k}{1-k^{2}} )=-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 $ .
BETA
