2021 AMC 12B Problem 18

Let $p=|z|^2=z\bar z$ and $s=z+\bar z.$ Then $|z+2|^2=p+2s+4,$ and $|z^2+1|^2$ $=p^2+(z^2+\bar z^2)+1$ $=p^2+(s^2-2p)+1.$ Substituting, $12p=2(p+2s+4)$ $+p^2+s^2-2p+1+31,$ which simplifies to $p^2-12p+s^2+4s+40=0.$ Completing the square gives $(p-6)^2+(s+2)^2=0,$ so $p=6$ and $s=-2.$ Then $z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.$ Thus, the correct answer is A.