Complex Numbers And Quadratic Equations question 100
Question: If for complex numbers $ z_1 $ and $ z_2 $ , $ arg(z_1/z_2)=0, $ then $ |z_1-z_2| $ is equal to
Options:
A) $ |z_1|+|z_2| $
B) $ |z_1|-|z_2| $
C) $ ||z_1|-|z_2|| $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ |z_1-z_2{{|}^{2}} $ $ =|z_1{{|}^{2}}+|z_2{{|}^{2}}-2|z_1||z_2|\cos ({\theta_1}-{\theta_2}) $ where $ {\theta_1}=arg(z_1) $ and $ {\theta_2}=arg(z_2) $ Since $ argz_1-argz_2=0 $ \ $ |z_1-z_2{{|}^{2}}=|z_1{{|}^{2}}+|z_2{{|}^{2}}-2|z_1||z_2| $ $ ={{(|z_1|-|z_2|)}^{2}} $
Þ $ |z_1-z_2|=||z_1|-|z_2|| $
BETA
