Question

$P$ represents the variable complex number $z$. Find the locus of $P$, if Im$\left[\frac{2z+i}{iz-1}\right]=-1$

Answer 1

Let $z=x+iy$
$\frac{2z+i}{iz-1}=\frac{2(x+iy)}{i(x+iy)-1}=\frac{2x+i(1+2y)}{-y-1+ix}$
$=\frac{2x+i(1+2y)\times -(1+y)=ix}{-(1+y)+ix\times -(1+y)-ix}$
Given, $Im=\left[\frac{2z+i}{iz-1}\right]=-1$
$\Rightarrow \frac{-2x^2-(1+2y)(1+y)}{{[-(1+y\left)\right]}^2+x^2}=1$
$\Rightarrow \frac{-2x^2-(1+y+2y+2y^2)}{1+y^2+2y+x^2}=-1$
$\Rightarrow -2x^2-1-3y-2y^2=-1-y^2-2y-x^2$
$\Rightarrow x^2+y^2+y=0$
$\therefore$ locus of the $P$ is therefore $x^2+y^2+y=0$.