Find the $z\in \Bbb C$ such that $\bar z=z^3$ Here is what I've done.  Let $z\in \Bbb C$ be of the form $a+ib$, lets assume in this case that $a,b\not=0$. We have that $\bar z=a-ib$, and $z^3=a^3-3ab^2+i(3a^2b-b^3)$, so to find the correct $z$, it has to satisfy: 
$$\begin{cases}
a= a^3-3ab^2\\
-b= 3a^2b-b^3\\
\end{cases}$$
wich is the same as: 
$$\begin{cases}
a=a( a^2-3b^2)\\
-b=b (3a^2-b^2)\\
\end{cases} \Rightarrow \begin{cases}
1=a^2-3b^2\\
1=b^2 -3a^2\\
\end{cases}$$
then, $a^2-3b^2=b^2 -3a^2 \Rightarrow 4a^2=4b^2 \Rightarrow a^2=b^2 \Rightarrow \begin{cases}
a=b \;\text{ or,}\\
a=-b\\
\end{cases}$
I thought this was correct, and to try it out I went to Wolfram, and it turns out that the examples that came to my mind didn't satisfied this. What did I do wrong?
 A: You should use radial coordinates. If $z \neq 0$, then your equation is equivalent to $|z|^2 = z^4$. 
For $r$, we have $r^2 = r^4$, which amounts to $r^2 - 1 = (r - 1)(r + 1) = 0$. Since $r > 0$ by hypothesis, we have $r = 1$.
Pick out polar and radial parts: if $z = r e^{i \theta}$, then we must have that $4 \theta$ is a multiple of $2 \pi$, hence $\theta = 0, \pi/2, \pi, 3 \pi/2$.
The amusing thing is that in assuming $a,b$ both nonzero, you excluded all the possible solutions, which are $1, i, -1, -i$, and $0$.
A: Note that $|\bar{z}|=|z|$. Thus, $|z|=|\bar{z}|=|z|^3\implies|z|\in\{0,1\}$.
If $|z|=0$, then $z=0$.
If $|z|=1$, then $\bar{z}=1/z$. Thus, $1/z=\bar{z}=z^3\implies z^4=1$. Thus, $z\in\{1,-1,i,-i\}$.
Just to be clear, there are $5$ solutions: $z\in\{0,1,-1,i,-i\}$

To answer Ana's comment: this generalizes to $\bar{z}=z^n$. $|z|=|\bar{z}|=|z|^n\implies|z|\in\{0,1\}$.
If $|z|=0$, then $z=0$.
If $|z|=1$, then $\bar{z}=1/z$. Thus, $1/z=\bar{z}=z^n\implies z^{n+1}=1$. That is,
$$
z\in\{0\}\cup\{e^{i2k\pi/(n+1)}:k\in\mathbb{Z}\text{ and }0\le k\le n\}
$$
These are precisely the solutions to $z^{n+2}-z=0$.
A: Ana, take a look at this:
Say we have $a+b=1$ and $a-b=1$ It follows very easily that $a=1$ AND $b=0$. However, with your approach, we could say $a+b=a-b$ and here we have infinite options for $a$
