Complex Numbers And Quadratic Equations question 135
Question: If $ z_1=a+ib $ and $ z_2=c+id $ are complex numbers such that $ |z_1|=|z_2|=1 $ and $ R(z_1\overline{z_2})=0, $ then the pair of complex numbers $ w_1=a+ic $ and $ w_2=b+id $ satisfies [IIT 1985]
Options:
A) $ |w_1|=1 $
B) $ |w_2|=1 $
C) $ R(w_1\overline{w_2})=0, $
D) All the above
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ |z_1|=|z_2|=1 $ , we have $ z_1=\cos {\theta_1}+i\sin {\theta_1},z_2=\cos {\theta_2}+i\sin {\theta_2} $ where $ {\theta_1}=arg(z_1) $ and $ {\theta_2}=arg(z_2) $ Also, $ z_1=a+ib $ and $ z_2=c+id. $ Therefore $ a=\cos {\theta_1} $ , $ b=\sin {\theta_1},c=\cos {\theta_2} $ and $ d=\sin {\theta_2} $ Also, $ R(z_1{{\overline{z}}_2})=0 $
Þ $ R[(\cos {\theta_1}+i\sin {\theta_1})(\cos {\theta_2}-i\sin {\theta_2})]=0 $
Þ $ R[(\cos ({\theta_1}-{\theta_2})+i\sin ({\theta_1}-{\theta_2})]=0 $
Þ $ \cos ({\theta_1}-{\theta_2})=0 $
Þ $ {\theta_1}-{\theta_2}=\frac{\pi }{2} $
Þ $ {\theta_1}={\theta_2}+\frac{\pi }{2} $ Now, $ w_1=a+ic=\cos {\theta_1}+i\cos {\theta_2} $ $ =\cos {\theta_1}+i\sin {\theta_1} $
Þ $ |w_1|=1 $ Similarly, $ |w_2|=1 $ Next $ w_1{{\overline{w}}_2}=(\cos {\theta_1}+i\sin {\theta_1})(\cos {\theta_2}-i\sin {\theta_2}) $ $ =\cos ({\theta_1}-{\theta_2})+i\sin ({\theta_1}-{\theta_2}) $
Þ $ |w_1{{\overline{w}}_2}|=1 $ Finally, $ R({{\overline{w}}_1}w_2)=R(w_2{{\overline{w}}_1}) $ $ =R[(\cos {\theta_2}+i\sin {\theta_2})(\cos {\theta_1}-i\sin {\theta_1})] $ $ =R[\cos ({\theta_2}-{\theta_1})+i\sin ({\theta_2}-{\theta_1})] $ $ =\cos ({\theta_2}-{\theta_1})=\cos ( \frac{-\pi }{2} )=0 $
BETA
