Complex Numbers And Quadratic Equations question 99
Question: For any two complex numbers $ z_1,z_2 $ we have $ |z_1+z_2{{|}^{2}}= $ $ |z_1{{|}^{2}}+|z_2{{|}^{2}} $ then
Options:
A) $ Re( \frac{z_1}{z_2} )=0 $
B) $ Im( \frac{z_1}{z_2} )=0 $
C) $ Re(z_1z_2)=0 $
D) $ Im(z_1z_2)=0 $
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Answer:
Correct Answer: A
Solution:
We have $ |z_1+z_2{{|}^{2}}=|z_1{{|}^{2}}+|z_2{{|}^{2}} $
Þ $ |z_1{{|}^{2}}+|z_2{{|}^{2}}+2|z_1||z_2|\cos ({\theta_1}-{\theta_2}) $ $ =|z_1{{|}^{2}}+|z_2{{|}^{2}} $ Where $ {\theta_1}=arg(z_1),{\theta_2}=arg(z_2) $
Þ $ \cos ({\theta_1}-{\theta_2})=0\Rightarrow {\theta_1}-{\theta_2}=\frac{\pi }{2} $
Þ $ arg( \frac{z_1}{z_2} )=\frac{\pi }{2}\Rightarrow Re( \frac{z_1}{z_2} )=\frac{|z_1|}{|z_2|}\cos ( \frac{\pi }{2} )=0 $ Note: Also $ Re( \frac{z_1}{z_2} )=0\Rightarrow Re(z_1\overline{z_2})=0 $
Þ $ z_1\overline{z_2} $ is purely imaginary.
BETA
