Question: How would you be able to solve this equation?

$$
z^2 \overline z = z
$$

for $z \in \mathbb{C}$?

Thanks in advance!

Answer: 
A trivial solution is $z=0$. Now suppose $z\neq 0$. Then since $\mathbb{C}$ is a field you may divide by $z$ and solve the easier equation
$$ z\bar{z}=1 $$
Therefore the solutions are $0$ and all complex numbers with norm $1$.
Let $$z=x+iy$$ then you have to solve $$(x+iy)^2(x-iy)-(x+iy)=0$$
It is equal to $$\left( x+iy \right) \left( {x}^{2}+{y}^{2}-1 \right) =0$$