HMMT November

Since $\omega^{3}=1$, and $\omega \neq 1, \omega$ is a third root of unity. For any complex number $z,|z|^{2}=z \cdot \bar{z}$. Letting $z=a+b \omega+c \omega^{2}$, we find that $\bar{z}=a+c \omega+b \omega^{2}$, and $$|z|^{2} =a^{2}+a b \omega+a c \omega^{2}+a b \omega^{2}+b^{2}+b c \omega+a c \omega+b c \omega^{2}+c^{2} =\left(a^{2}+b^{2}+c^{2}\right)+(a b+b c+c a)(\omega)+(a b+b c+c a)\left(\omega^{2}\right) =\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a) =\frac{1}{2}\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)$$ where we have used the fact that $\omega^{3}=1$ and that $\omega+\omega^{2}=-1$. This quantity is minimized when $a, b$, and $c$ are as close to each other as possible, making $a=3, b=4, c=5$ the optimal choice, giving $|z|^{2}=3$. (A smaller value of $|z|$ requires two of $a, b, c$ to be equal and the third differing from them by at most 2, which is impossible.) So $|z|_{\min }=\sqrt{3}$.