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.
 A: A complex number $w$ is real if and only if it equals its conjugate $\bar{w}$. So you need to solve
$$
\frac{z+2}{z(z+1)}=\frac{\bar{z}+2}{\bar{z}(\bar{z}+1)}
$$
Cross multiplication gives
$$
z\bar{z}^2+z\bar{z}+2\bar{z}^2+2\bar{z}=z^2\bar{z}+z\bar{z}+2z^2+2z
$$
and we can transfer everything to the right-hand side:
$$
z\bar{z}(z-\bar{z})+2(z^2-\bar{z}^2)+2(z-\bar{z})=0
$$
This has the entire real line ($z=\bar{z}$) as a solution, points $z=0$ and $z=-1$ excluded, of course; removing this factor we find all other solutions:
$$
z\bar{z}+2(z+\bar{z})+2=0 \tag{*}
$$
Now we can substitute $z=x+yi$:
$$
x^2+y^2+4x+2=0
$$
that is,
$$
(x+2)^2+y^2=2
$$
which is a circle.
You can avoid passing to real coordinates by noticing that the equation (*) can be rewritten as $z\bar{z}+2z+2\bar{z}+4=2$, so $(z+2)(\bar{z}+2)=2$ and finally
$$
|z+2|^2=2
$$
which is clearly a circle with center at $2$ and radius $\sqrt{2}$.
A: HINT: Multiply the fraction by 
$$
\frac{\bar z(\bar z-1)}{\bar z(\bar z-1)}\;.
$$
In this way the denominator will be real and you should only look at the numerator, splitting it into its real and imaginary part.
A: A little Partial Fraction Decomposition before rationalization of the denominator will reduce calculation. 
$$\dfrac{z+2}{z(z+1)}=\dfrac1{z+1}+\dfrac{2(z+1-z)}{z(z+1)}=\dfrac2z-\dfrac1{z+1}$$
Now the imaginary part of $\dfrac2z$ is $-\dfrac{2y}{x^2+y^2}$
and that of $\dfrac1{z+1}$ is $-\dfrac y{(x+1)^2+y^2}$
Finally, $\dfrac{z+2}{z(z+1)}$ will be real if $-\dfrac{2y}{x^2+y^2}=-\dfrac y{(x+1)^2+y^2}$
$\iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y\{(x+2)^2+y^2-2\}$
A: 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.
