<s>[INST] If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal. Let $z=a+b\omega + c\omega^2$
$$z=a+b\omega -c (1+\omega)$$
$$z=a-c+\omega (b-c)$$
Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$
How should I proceed?
 [/INST] Note that:
$$|a+b\omega+c{\omega}^2|^2=(a+b\omega+c{\omega}^2)(a+b{\omega}^2+c\omega)$$
This is because $\bar{\omega}={\omega}^2$, and $\bar{{\omega}^2}=\omega$. If you multiply out the terms, you'll get $a^2+b^2+c^2-ab-bc-ca$, which is equal to $\frac {(a-b)^2+(b-c)^2+(c-a)^2}{2}$. Since $a,b,c$ are not all equal, two or all three of these square terms cannot be $0$. Hence for minimum let any one of them be $0$ and let the remaining variable differ by $1$. You will get final answer $1$.
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