The minimum value of $|a + b\omega + c\omega^2|$, where $a$, $b$ and $c$ are all not equal integers and $\omega$ ($\neq 1$) is a cube root of unity, is
Step-by-Step Solution
Key Concept: The identity $|a+b\omega+c\omega^2|^2 = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$ reduces the problem to minimising a sum of squared differences over distinct integers.
**Step 1: Use the modulus-squared identity**
$|a+b\omega+c\omega^2|^2 = a^2+b^2+c^2-ab-bc-ca = \dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$.
**Step 2: Minimise over distinct integers**
Since $a,b,c$ are pairwise unequal integers, the minimum of $(a-b)^2+(b-c)^2+(c-a)^2$ is achieved with consecutive integers, e.g. $\{0,1,2\}$: value $= 1+1+4=6$.
**Step 3: Conclude**
$|a+b\omega+c\omega^2|^2_{\min} = \dfrac{6}{2} = 3$, so the minimum modulus is $\sqrt{3}$.
Correct Answer: 1