Question

Let $\omega \ne 1$ be a cube root of unity. Then the minimum of the set
$\{|a+b\omega +c{\omega }^2{|}^2:a,b,c$ distinct non-zero integers} equals

Answer 1

$\{|a+b\omega +c{\omega }^2{|}^2$
$=(a+b\omega +c{\omega }^2)\overline{(a+b\omega +c{\omega }^2)}$
$=(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )$
$=(a^2+ab{\omega }^2+ac\omega +ab\omega +b^2{\omega }^3+bc{\omega }^2+ac{\omega }^2+bc{\omega }^4+c^2{\omega }^3)$
$=(a^2+{\omega }^3(b^2+c^2)+ab({\omega }^2+\omega )+ac({\omega }^2+\omega )+bc({\omega }^2+{\omega }^4\left)\right)$
$=a^2+b^2+c^2-ab-ac-bc$
$=\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$
$Leta>b>c\Rightarrow |a–b|\ge 1,|b–c|\ge 1,|a–c|\ge 2$
$\ge \frac{1}{2}[1+1+4]$
so, minimum of set $\{|a+b\omega +c{\omega }^2{|}^2$ $=3$