Complex Numbers And Quadratic Equations question 101
Question: If $ |z_1+z_2|=|z_1-z_2| $ , then the difference in the amplitudes of $ z_1 $ and $ z_2 $ is [EAMCET 1985]
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{2} $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
Squaring the given relations implies that $ x_1x_2+y_1y_2=0 $ Now $ ampz_1-ampz_2={{\tan }^{-1}}\frac{y_1}{x_1}-{{\tan }^{-1}}\frac{y_2}{x_2} $ $ ={{\tan }^{-1}}\frac{\frac{y_1}{x_1}-\frac{y_2}{x_2}}{1+\frac{y_1y_2}{x_1x_2}}={{\tan }^{-1}}\frac{y_1x_2-y_2x_1}{x_1x_2+y_1y_2} $ $ ={{\tan }^{-1}}\infty =\frac{\pi }{2} $ .
BETA
