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 $1$.
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Bumbble Comm
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Let $$z=x+iy$$ then you have to solve $$(x+iy)^2(x-iy)-(x+iy)=0$$
It is equal to $$\left( x+iy \right) \left( {x}^{2}+{y}^{2}-1 \right) =0$$
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 $1$.