<s>[INST] 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?
 [/INST] 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 $a$
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