$\begin{array}{l}\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}=\frac{2x+1+2iy}{(1-y)+ix}\\ =\frac{\left[\right(2x+1)+2iy]\left[\right(1-y)-ix]}{(1-y{)}^2+x^2}\end{array}$

$\frac{(2x+1)(1-y)+2xy+i[-x(2x+1)+2y(1-y\left)\right]}{(1-y{)}^2+x^2}$

Since $\mathrm{Im}\left(\frac{2z+1}{iz+1}\right)=-2$, we have

$\frac{-x(2x+1)+2y(1-y)}{(1-y{)}^2+x^2}=-2$

or $-2x^2-2y^2-x+2y=-2\left(1+y^2-2y\right)-2x^2$

or $x+2y-2=0$

which is a straight line being of 1st degree.