Complex Numbers And Quadratic Equations question 688
Question: If z, $ \omega z $ ane $ \bar{\omega }z $ are the vertices of a triangle, then the area of the triangle will be (where $ \omega $ is cube root of unity):
Options:
A) $ \frac{3|z{{|}^{2}}}{2} $
B) $ \frac{3\sqrt{3}|z{{|}^{2}}}{2} $
C) $ \frac{\sqrt{3}|z{{|}^{2}}}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let the point A represents the complex number z, B represents $ \omega z $ and C represents $ \bar{\omega }z $ . $ \omega \And \bar{\omega } $ are complex cube roots of unity clearly coz means rotation of z by $ \frac{2\pi }{3}and{{\omega }^{2}}z $ $ (=\bar{\omega }z) $ means rotation of $ \omega zby\frac{2\pi }{3} $
$ \therefore \angle AOB = \angle BOC = \angle COA =\frac{2\pi }{3} $ also $ OA= OB = OC = | z | $ . That is the $ \Delta ABC $ is equilateral. Now $ AC = 2AD = 2 ( OA cos 30{}^\circ ) $ $ =2|z|\frac{\sqrt{3}}{2}=\sqrt{3}| z | $ $ Area of \Delta ABC= \frac{\sqrt{3}}{2}{{( side )}^{2}} = \frac{3\sqrt{3}}{2} {{| z |}^{2}} $
BETA
