, 2017

Solution:

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
$$\begin{aligned} |z|^{2} & = a^{2} + a b \omega + a c \omega^{2} + a b \omega^{2} + b^{2} + b c \omega + a c \omega + b c \omega^{2} + c^{2} \\ & = \left(a^{2} + b^{2} + c^{2}\right) + (a b + b c + c a)(\omega) + (a b + b c + c a)\left(\omega^{2}\right) \\ & = \left(a^{2} + b^{2} + c^{2}\right) - (a b + b c + c a) \\ & = \frac{1}{2}\left((a-b)^{2} + (b-c)^{2} + (c-a)^{2}\right), \end{aligned}$$
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 $|z|_{\text{min}} = \sqrt{3}$.