<s>[INST] Complex roots of $z^3 + \bar{z} = 0$ 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.
 [/INST] EDIT in view of the comments bellow by JimConant and PeterTaylor. If there is still any error the fault is mine.
This is an alternative solution to the trigonometric one. We will use the algebraic method. Let $z=x+iy$. We have
$$\begin{eqnarray*}
0 &=&z^{3}+\overline{z} \\
0 &=&\left( x+iy\right) ^{3}+\left( x-iy\right)  \\
&=&x^{3}+x-3xy^{2}+i\left( 3x^{2}y-y^{3}-y\right)  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
0=x(x^{2}+1-3y^{2}) \\ 
0=y(y^{2}+1-3x^{2}).
\end{array}
\right. 
\end{eqnarray*}\tag{1}$$
One of the roots is $$x_{1}=y_{1}=0.\tag{1a}$$ The remaining real roots satisfy the system
$$\begin{eqnarray*}
\left\{ 
\begin{array}{c}
0=x^{2}+1-3y^{2} \\ 
0=y^{2}+1-3x^{2}
\end{array}
\right.  &\Leftrightarrow &\left\{ 
\begin{array}{c}
0=x^{2}+1-3y^{2} \\ 
0=\frac{1}{3}+\frac{1}{3}x^{2}+1-3x^{2}
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
0=x^{2}+1-3y^{2} \\ 
0=4-8x^{2}
\end{array}
\right.  \\
&\Leftrightarrow &\left\{ 
\begin{array}{c}
0=1-2y^{2} \\ 
0=1-2x^{2}.
\end{array}
\right. 
\end{eqnarray*}
\tag{2}$$
The last system means that $$y=\pm x\tag{3}$$ and that 
$$\begin{eqnarray*}
x &=&\pm\frac{1}{2}\sqrt{2}, \\
y &=&\pm\frac{1}{2}\sqrt{2}.
\end{eqnarray*}\tag{3a}$$
Combining the above results, we conclude that the following five complex numbers
$$
z_{1} =0,\tag{4}$$
$$
z_{2} =\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2},\quad z_{3} =-\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2},
\tag{5}$$
$$z_{4} =\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2},\quad z_{5} =-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2},
\tag{6}$$
are the solutions of the given equation $$z^{3}+\overline{z}=0.\tag{7}$$
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