question_answer

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):               

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

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 \[\omega z\,\,by\,\,\frac{2\pi }{3}\] \[\therefore  \,\,\angle AOB = \angle BOC = \angle COA =\frac{2\pi }{3}\] also \[\operatorname{OA}= OB = OC = \left| z \right|\]. That is the \[\Delta \,ABC\] is equilateral. Now \[\operatorname{AC} = 2AD = 2 \left( OA cos 30{}^\circ  \right)\] \[=2\,|z|\frac{\sqrt{3}}{2}\,\,=\,\,\sqrt{3}\,\left| \,z\, \right|\] \[\operatorname{Area} \,of \,\Delta ABC= \frac{\sqrt{3}}{2}{{\left( side \right)}^{2}} = \frac{3\sqrt{3}}{2} {{\left| \,z\, \right|}^{2}}\]