Find the geometric locus $z \in \mathbb C$ so that $\frac{z+2}{z(z+1)}\in \mathbb R$

Question

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.

Answer 1

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.

Answer 2

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 $\sqrt{2}$.

Answer 3

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

$\iff0=y(2x^2+2y^2+4x+2-x^2-y^2)=y\{(x+2)^2+y^2-2\}$

Answer 4

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.