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 $z=0,1,-1,i,-i$.
\Rightarrow$
or $\implies$$\implies$
. It definitely looks better than $=>$. However, in this case writing simply $=$ might be better. $\endgroup$