Discussion Forum

If the imaginary part of $\frac{2z+1}{iz+1}$ is -2  , then the locus of the point representing  z in the complex plane is 

Answer:

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