Complex Numbers And Quadratic Equations question 119
Question: If z and $ \omega $ are two non-zero complex numbers such that $ |z\omega |=1 $ and $ arg(z)-arg(\omega )=\frac{\pi }{2}, $ then $ \bar{z}\omega $ is equal to [AIEEE 2003]
Options:
A) 1
B) - 1
C) i
D) - i
Show Answer
Answer:
Correct Answer: D
Solution:
$ |z||\omega |=1 $ ……(i) and $ arg( \frac{z}{\omega } )=\frac{\pi }{2}\Rightarrow \frac{z}{\omega }=i $
Þ $ | \frac{z}{\omega } |=1 $ …..(ii) From equation (i) and (ii) $ |z|=|\omega |=1 $ and $ \frac{z}{\omega }+\frac{{\bar{z}}}{{\bar{\omega }}}=0;z\bar{\omega }+\bar{z}\omega =0 $ $ \bar{z}\omega =-z\bar{\omega }=\frac{-z}{\omega }\bar{\omega }\omega $ ; $ \bar{z}\omega =-i|\omega {{|}^{2}}=-i. $ .
BETA
