The complex numbers `z_1 z_2 and z3` satisfying `(z_1-z_3)/(z_2-z_3) =(1- i sqrt(3))/2` are the vertices of triangle which is

Options

A. of area zero
B. right angled isosceles
C. equilateral
D. obtuse angled isosceles

Answer

C

Explanation

`(z_1-z_3)/(z_2-z_3)=(1-isqrt(3))/(2)=((1-isqrt(3))(1 + i sqrt(3)))/(2(1+isqrt(3)))`
`=(1-i^23)/(2(1+isqrt(3)))`
`= (4)/(2(1+isqrt(3)))`
`(2)/(2(1+isqrt(3)))`
`rArr (z_2-z_3)/(z_1-z_3)=(1+sqrt(3))/2= cos ""pi/3+ i sin""pi/3`
`rArr |(z_1-z_3)/(z_1-z_3)|=1 and arg ((z_2 -z_2)/(z_1-z_3))=pi/3`
Hence the triangle is an equilateral .
Alternate solution
`therefore (z_1-z_3)/(z_2-z_3)=(1-isqrt3)/(2)`
`rArr ((z_2-z_1)/(z_1-z_2))= pi/3 and also |(z_2-z_3)/(z_1-z_3)|=1`
Therefore , triangle is equilateral .