Complex Numbers And Quadratic Equations question 453
Question: If for complex numbers, $ z_1 $ and $ z_2 $ arg ( $ z_1 $ ) - arg( $ 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 | $
0
Show Answer
Answer:
Correct Answer: C
Solution:
[c] 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) $ .given, Arg $ (z_1-z_2) $ =0.
$ \Rightarrow {{| z_1-z_2 |}^{2}}={{| z_1 |}^{2}}+{{| z_2 |}^{2}}-2\operatorname{Re}(z_1 \overline{z_2}) $ $ ={{(| z_1 |-| z_2 |)}^{2}} $ $ \Rightarrow | z_1 - z_2 | = | | z_1 | - | z_2 | | $
BETA
