Question

a, b, c are integers, not all simultaneously equal, and $\omega$ is cube root of unity $(\omega \ne 1)$, then the minimum value of $|a+b\omega +c{\omega }^2|$ is

• A. $0$
• B. $1$
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $1$

$|a+b\omega +c{\omega }^2|$

We know $1+\omega +{\omega }^2=0$

$\Rightarrow {\omega }^2=-(1+\omega )$

$\Rightarrow |a+b\omega +c{\omega }^2|=|a+b\omega +c(-(1+\omega ))|$

$=|(a-c)+(b-c)|\ge |a-c|+\left|(b-c)\omega \right|$

$\Rightarrow \left|\omega \right|=1$

$\Rightarrow |a-c|+|b-c|$

Let $a-c=0,$ $b-c=0$ for minimum.

$\Rightarrow a=c$ $b=c$ which is contradictory for question.

$\therefore$ for $a=c=0,$ $|b-c|=1$

$\Rightarrow b=1or(-1)$

$\therefore$ Minimum value is $1.$