Complex Numbers And Quadratic Equations question 108
Question: $ |z_1+z_2|=|z_1|+|z_2| $ is possible if [MP PET 1999; Pb. CET 2002]
Options:
A) $ z_2={{\overline{z}}_1} $
B) $ z_2=\frac{1}{z_1} $
C) $ arg(z_1)= $ arg $ (z_2) $
D) $ |z_1|=|z_2| $
Show Answer
Answer:
Correct Answer: C
Solution:
$ |z_1+z_2|=|z_1|+|z_2| $ $ |z_1+z_2{{|}^{2}}=|z_1{{|}^{2}}+|z_2{{|}^{2}}+2|z_1||z_2| $
Þ $ |z_1{{|}^{2}}+|z_2{{|}^{2}}+2Re|z_1{{\bar{z}}_2}| $ $ =|z_1{{|}^{2}}+|z_2{{|}^{2}}+2|z_1||z_2| $
Þ $ 2Re|z_1{{\bar{z}}_2}|=2|z_1||z_2| $
Þ $ 2|z_1||{{\bar{z}}_2}|\cos ({\theta_1}-{\theta_2})=2|z_1||z_2| $
Þ $ \cos ({\theta_1}-{\theta_2})=1 $ or $ {\theta_1}-{\theta_2}=0 $ \ $ arg(z_1)=arg(z_2) $
BETA
