Answer:
Option B
Explanation:
Let z=x+iy
$\therefore$ $\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}=\frac{2x+1+iy}{(-y+1)+ix}$
=$\frac{[(2x+1)+iy][(-y+1)-ix]}{[(-y+1)+ix][(-y+1)-ix]}$
=$\frac{(2x+1)(-y+1)-(2x+1)xi+y(-y+1)j+xy}{(-y+1)^{2}+x^{2}}$
=$\frac{(2x+1)(-y+1)+xy+(-y^{2}+y-2x^{2}-x)i}{(-y+1)^{2}+x^{2}}$
It is given that imaginary part of $\frac{2z+1}{iz+1}=-2$
$\therefore$ $\frac{-y^{2}+y-2x^{2}-x}{(-y+1)^{2}+x^{2}}=-2$
$\Rightarrow$ $-y^{2}+y-2x^{2}-x=-2(-y+1)^{2}-2x^{2}$
$\Rightarrow$ $-y^{2}+y-x=-2y^{2}-2+4y$
$\Rightarrow$ $y^{2}-3y-x+2=0$
Which represent the equations of parabola
