Harvard-MIT Mathematics Tournament

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 $\left(z^{2} + 2\right)^{2} - z + 2 = z^{4} + 4z^{2} - z + 6$. Factoring it out, we obtain $\left(z^{2} - z + 2\right)\left(z^{2} + z + 3\right) = z^{4} + 4z^{2} - z + 6$. Finally, the second term leads to the solutions $z = \frac{-1 \pm i \sqrt{11}}{2}$.