Question

If ${{I}_{m}}\left( \frac{z-1}{2z+1} \right)=-4,$ then locus of z is

• A. ellipse
• B. parabola
• C. straight line
• D. circle

Answer 1

Answer: (D)

circle

Answer 2

Let $z=x+i y$
$\therefore \frac{z-1}{2 z+1}=\frac{x+i y-1}{2 x+2 i y+1}$
$=\frac{(x-1)+i y}{(2 x+1)+2 i y} \times \frac{(2 x+1)-2 i y}{(2 x+1)-2 i y}$
$=\frac{(x-1)(2 x+1)-2 i y(x-1)+i y(2 x+1)+2 y^{2}}{(2 x+1)^{2}+4 y^{2}}$
=\frac{\left\\{(x-1)(2 x+1)+2 y^{2}\right\\}+i y\\{-2 x+2+2 x+1\\}}{(2 x+1)^{2}+4 y^{2}}
According to question
$I_{m}\left(\frac{z-1}{2 z+1}\right)=-4$
$\therefore \frac{3 y}{(2 x+1)^{2}+4 y^{2}}=-4$
$\Rightarrow-\frac{3 y}{4}=4 x^{2}+4 y^{2}+4 x+1$
$\Rightarrow 16 x^{2}+16 y^{2}+16 x+3 y+4=0$
This equation represents a circle.
$\therefore$ The locus of $z$ is a circle.