Complex Numbers And Quadratic Equations question 83
Question: If $ z_1 $ and $ z_2 $ are two non-zero complex numbers such that $ |z_1+z_2|=|z_1|+|z_2|, $ then arg $ (z_1)- $ arg $ (z_2) $ is equal to [IIT 1979, 1987; EAMCET 1986; RPET 1997; MP PET 2001; AIEEE 2005]
Options:
A) $ -\pi $
B) $ -\frac{\pi }{2} $
C) $ \frac{\pi }{2} $
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ z_1=r_1 $ $ (\cos {\theta_1}+i\sin {\theta_1}) $ , $ z_2=r_2 $ $ (\cos {\theta_2}+i\sin {\theta_2}) $
$ \therefore $ $ |z_1+z_2|=[{{(r_1\cos {\theta_1}+r_2\cos {\theta_2})}^{2}} $ $ +{{(r_2\sin {\theta_1}+r_2\sin {\theta_2})}^{2}}{{]}^{1/2}} $ $ ={{[r_1^{2}+r_2^{2}+2r_1r_2\cos ({\theta_1}-{\theta_2})]}^{1/2}}={{[{{(r_1+r_2)}^{2}}]}^{1/2}} $ $ arg(z_2)=\theta $ Therefore $ \cos ({\theta_1}-{\theta_2})=1\Rightarrow {\theta_1}-{\theta_2}=0\Rightarrow {\theta_1}={\theta_2} $ Thus arg $ (z_1)-arg(z_2)=0 $ . Trick: $ |z_1+z_2|=|z_1|+|z_2|\Rightarrow z_1,z_2 $ lies on same straight line.
$ \therefore argz_1=argz_2\Rightarrow argz_1-argz_2=0 $
BETA
