Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$ 
Find the geometric locus of the set of $z \in \mathbb C$ so that 
  $$\frac{z+2}{z(z+1)}\in \mathbb R$$
Source: IME (Military Engineering Institute, Brazil, entrance exam, 1974)

My attempt: With the notation $z=a+bi$, the solution provided in a book is either the line $b=0$ or the circle $(a+2)^2+b^2=2$  but I could not find it, or not able to recognize this locus set from the algebraic development I did (or the solution or the statement has some mistake).  
Hints and solutions are welcomed.

Using
$$
z = \sqrt{x^2+y^2}e^{\arctan\frac yx}
$$
we have
$$
\frac{z+2}{z(z+1)} = \frac{\rho_1 e^{i\phi_1}}{\rho_2 e^{i\phi_2}\rho_3 e^{i\phi_3}}
$$
and we seek for
$$
\arctan\frac{y}{x+2}-\arctan\frac yx -\arctan\frac{y}{x+1} = 0
$$
then
$$
\tan\left(\arctan\frac{y}{x+2} -\arctan\frac{y}{x+1}\right) = \frac yx
$$
or
$$
y(y^2+x^2+4x+2) = 0
$$
NOTE
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b}
$$
so
$$
\tan\left(\arctan\frac{y}{x+2} -\arctan\frac{y}{x+1}\right) = \frac{y}{y^2+x^2+3x+2}
$$
etc.
