If `Imz((z-1)/(2z+1))=-4`, then locus of z is

Options

A. an ellipse
B. a parabola
C. a straight line
D. a circle

Answer

D

Explanation

Let `z=x+iy`. Then,
`(z-1)/(2z+1) = (((x-1)+iy)(2x+1)-2iy)/(((2x+1)+2iy)(2x+1)-2iy)`
`rArr (z-1)/(2z+1) = (x-1)+(2x+1)+2y^(2)+i(y(2x+1)-2y(x-1))/((2x+1)^(2)+4y^(2))`
`"Im"(z-1)/(2z+1)=(3y)/((2x+1)^(2)+4y^(2))`
Now, `"Im" (z-1)/(2z+1)=-4`
`rArr (3y)/((2x+1)^(2)+4y^(2))=-4 rArr x^(2)+y^(2)+x+3/16y+1/4=0`
Clearly, it represents a circle.