Question on complex numbers and cube root of unity

Question

If $a$, $b$, $c$ are distinct integers and $\omega$ is a cube root of unity then minimum value of $|a + b\omega + c\omega^2| + |a+b\omega^2 + c\omega|$ is?

Now, I know the identity $a^3 + b^3 +c^3 -3abc=(a+b+c)(a + b\omega + c\omega^2)(a+b\omega^2 + c\omega)$.

I don't know how to apply this to the question. Or is there something else that I am missing?

Any help would be appreciated.

Answer 1

Let

$$A=|a + b\omega + c\omega^2|$$

$$B=|a+b\omega^2 + c\omega|$$

note that $B=\bar A=A\implies A+B=2A$

then

$$A+B=2A=2\left|a + b\left(-\frac12+i\frac{\sqrt3}{2} \right) + c\left(-\frac12-i\frac{\sqrt3}{2} \right)\right|=2\left|a -\frac12b-\frac12c+i\frac{\sqrt3}{2}\left(b-c\right)\right|=2\left(a^2+\frac14b^2+\frac14c^2-ab-ac+\frac12bc+\frac34b^2+\frac34c^2-\frac32bc\right)^{\frac12}=2\left(a^2+b^2+c^2-ab-bc-ca\right)^{\frac12}=2\left(\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\right)^{\frac12}$$

For $a=1,b=2,c=3$ we obtain $A+B=2\sqrt 3$.

Answer 2

Hint: $|a+b\omega+c\omega^2|=|\overline{a+b\omega+c\omega^2}| = |a+b\bar\omega+c\bar\omega^2| = |a+b\omega^2+c\omega|$.