<s>[INST] 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.
 [/INST] 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 $-\dfrac{2y}{x^2+y^2}=-\dfrac y{(x+1)^2+y^2}$
$\iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y\{(x+2)^2+y^2-2\}$
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