Minimizing $|a+bw+cw^2|$ where $a,b,c\in\mathbb Z$ and $w = \zeta_3$. 
If $a,b,c\in\mathbb Z$ are not all equal and $w$ is a cube root of unity $(w\neq 1$), then the minimum value of:
  $$|a+bw+cw^2|$$
  is what?

I'm pretty stuck with the above problem.  Could someone help me out?
 A: Geometry! $1$, $w$ and $w^2$ are corners of an equilateral triangle centered at 0. Seen as vectors, they are moves in 3 directions in space, along a triangular lattice. So just plot a triangular grid and ask yourself what the absolute value tells you. In the image below, increasing $a$ by one moves in the right direction by 1, and increasing $b$ by 1 goes in the direction of the other red arrow ($c$ goes in the third direction which doesn't have an arrow but you see where this is going). If $a=b=c$, you get back to the origin. As the rules say they can't all be equal, you have to modify at least 1, so you end up at least 1 step away from the origin. If all 3 have to be different, you end up even further.

A: I have solved it and given you some hint . 
$\left|a+b\omega+c\omega^2\right|=\sqrt{\left(a-\frac{b}{2}-\frac{c}{2}\right)^2+\frac{3}{4}(c-b)^2}=\sqrt{\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)}$
This is minimum when $a=b$ and $(b-c)^2=(c-a)^2=1\implies$ The minimum value is $1$.
A: Using $|z|^2 = z\bar{z}$
and Using $\displaystyle \bar{\omega} = \omega^2$ and $\bar{\omega^2} = \omega$ and $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$
So $$|a+b\omega+c\omega^2|^2 = (a+b\omega+c\omega^2)\cdot \bar{(a+b\omega+c\bar{\omega^2})} = (a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
So we get $$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2+ab(\omega+\omega^2)+bc(\omega+\omega^2)+ca(\omega+\omega^2)$$
So $$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$$
Now Given $a,b,c$ not all are equal integers. So for $\min$ of right side expression 
Let $a=n\;,b=n\;,c=n+1\;,$ Where $n\in \mathbb{Z}$
So $$\min|a+b\omega+c\omega^2|^2 = \frac{1}{2}\left[1^2+1^2\right] = 1$$
So $$\min |a+b\omega+c\omega^2| = 1$$
