Solving complex numbers equation $z^3 = \overline{z} $ We have the following equation:
$$z^3 = \overline{z} $$
I set z to be $z = a + ib$ and since I know that $ \overline{z} = a - ib$. I was trying to solve it by opening the left side of the equation.
$$ z^3 = (a+ib)^3 \Rightarrow $$
$$ [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow $$
$$ a^3 - b^2a - 2b^2a + i (2a^2b + b^2a - b^3) $$
but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it.

You have a good start. Rewrite equation as $z^3-\bar{z} =0$, now do $z=a+bi$, so we get $$a^3-3b^2a-a+i(3a^2b-b^3-b) = 0$$
Now both the imaginary and the real part must be equal to zero, so we get the following system of equations $$a^3-3b^2a-a=0 \wedge 3a^2b-b^3-b=0$$
Factoring gives: 
$$a(a^2-3b^2-1)=0 \wedge b(3a^2-b^2-1)=0$$
So we have four possibilities: 


*

*$a=0, b=0$

*$a=0, 3a^2-b^2-1=0$

*$a^2-3b^2-1=0, b=0$

*$a^2-3b^2-1=0, 3a^2-b^2-1=0$



First one clearly gives $z=0$.

Second one: Substitute $a=0$ in to get $b^2-1=0$, so $b=1$ or $b=-1$. 
This gives $z=i$ and $z=-i$.

Third one: Substitute $b=0$ in to get $a^2-1=0$, so $a=1$ or $a=-1$. 
This gives $z=1$ and $z=-1$.

Fourth one: Subtract the first equation trice form the second. This gives $8b^2+2=0$, so $b^2+\frac{1}{4}=0$, so $b=\pm\frac{1}{2}i$. This gives no solutions, since we defined $b = \Im(z)$ and it must be real. 

Conclusion: The solutions are $z=0,1,-1,i,-i$. 
