TITLE: Complex roots of $z^3 + \bar{z} = 0$
QUESTION [12 upvotes]: I'm trying to find the complex roots of $z^3 + \bar{z} = 0$ using De Moivre.
Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know which root was the extra ).
I noticed that $z=a+bi$ and there exists $\theta$ such that the trigonometric representation of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$ .
It seems that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$
However, my trig is pretty rusty and I'm not quite sure where to go from here.

REPLY [2 votes]: That is a comment to the comment "should there be only 3 roots?"  in Zarrax answer.  Actually, the questions is "why isn´t there 9 roots?".  This is the right question to ask since the intersection of the two curves in $\mathbb{R}^2$
$$ x^3-3xy^2 +x = 0$$
$$ 3x^2y-y^3 -y = 0$$
is your solution set (just expand out $z^3+\bar{z}$).  Now, the intersection of two degree 3 equations should have $3x3= 9$ solutions (by Bezout´s theorem). The 4 roots we are missing at "infinity" or in the $\mathbb{C}^2$ plane. We could apply the same thing to the function $z^2+z$.  As complex polynomial, we should expect 2 roots.  As the real system $(x,y) \rightarrow (x^2-y^2+x,2xy+y)$ we expect four real roots.