8. Solve for all complex numbers $z$ such that $z^{4}+4 z^{2}+6=z$.
Answer: $\frac{1 \pm i \sqrt{7}}{2}, \frac{-1 \pm i \sqrt{11}}{2}$
Solution: Rewrite the given equation as $\left(z^{2}+2\right)^{2}+2=z$. Observe that a solution to $z^{2}+2=z$ is a solution of the quartic by substitution of the left hand side into itself. This gives $z=\frac{1 \pm i \sqrt{7}}{2}$. But now, we know that $z^{2}-z+2$ divides into $\left(z^{2}+2\right)^{2}-z+2=$ $z^{4}+4 z^{2}-z+6$. Factoring it out, we obtain $\left(z^{2}-z+2\right)\left(z^{2}+z+3\right)=z^{4}+4 z^{2}-z+6$. Finally, the second term leads to the solutions $z=\frac{-1 \pm i \sqrt{11}}{2}$.