Complex Numbers And Quadratic Equations question 669
Question: Let Z and W be two complex numbers such that $ | Z |\le 1, $ $ | W |\le 1 $ and $ | Z+iW |=| Z-i\overline{W} |=2. $ Then Z equals
Options:
A) 1 or i
B) i or $ -i $
C) 1 or $ -i $
D) i or $ -1 $
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ 2=| z+i\omega | \le | z |+| \omega | $ … (i)
$ \therefore ~| z |+| \omega |\ge 2 $ But given that $ | z | \le 1 and | \omega | \le 1 $ … (ii)
$ \Rightarrow | z |+| \omega |\le 2 $ From (1) and (2) $ | z |=| \omega |=1 $ Also $ |z+i\omega |=|z-i\overline{\omega }|\Rightarrow {{| z+i\omega |}^{2}}{{| z-i\overline{\omega } |}^{2}} $
$ \Rightarrow ( z+i\omega ) ( \bar{z}-i\bar{\omega } )=( \overline{z} +i \omega )( z-i\overline{\omega } ) $
$ \Rightarrow z\bar{z}+i\omega \bar{z}-iz\bar{\omega }+\omega \bar{\omega }=z\bar{z}-i\bar{z}\bar{\omega }+i\omega z+\omega \bar{\omega } $
$ \Rightarrow \omega \bar{z}-\bar{\omega }z+\bar{\omega }\bar{z}-\omega z=0 $ $ =(\omega +\bar{\omega })(\bar{z}-z)=0 $
$ \Rightarrow z=\bar{z}or\omega =-\bar{\omega }\Rightarrow {I_{m}}(z)=0\Rightarrow Re( \omega )=0 $ Also $ | z |=1,| \omega |=1\Rightarrow z=1or-1and\omega =i $ or $ -i $
BETA
