Question

Locate the complex numbers $z=x+iy$ such that.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 a straight line $x+2y-2=0$.

Answer 1

$=\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}=\frac{2x+1+2iy}{(1-y)+ix}$
$=\frac{[(2x+1)+2iy][(1-y)-ix]}{{(1-y)}^2+x^2}$
$=\frac{(2x+1)(1-y)+2xy+i[-x(2x+1)+2y(1-y)]}{{(1-y)}^2+x^2}$
Since $I_m\left(\frac{2z+1}{iz+1}\right)=-2$, we have
$\frac{-x(2x+1)+2y(1-y)}{{(1-y)}^2+x^2}=-2$
or $-2x^2-2y^2-x+2y=-2(1+y^2-2y)-2x^2$
or $x+2y-2=0$.