Question

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

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

Answer 1

The correct option is A ($1$)

Given that $a,b,c$ are integers not all equal, $\omega$ is cube root of unity $\ne$ 1
So, $1+\omega +{\omega }^2=0$
$\Rightarrow {\omega }^2=-(1+\omega )$
Consider $|a+b\omega +c{\omega }^2|$
Substitute ${\omega }^2=-(1+\omega )$, we get
$|a+b\omega +c{\omega }^2|=|a+b\omega +c[-(1+\omega )\left]\right|$
$=a+b\omega -c-c\omega$
$=a-c+(b-c)\omega$
Since $\omega =\frac{1\pm i\sqrt{3}}{2}$ as $\omega \ne 1$
$\Rightarrow (|a+b\omega +c{\omega }^2|=|a-c+(b-c\left)\right[\frac{1\pm i\sqrt{3}}{2}\left]\right|$
$=\left|\right[(a-c)+\frac{b-c}{2}]\pm i(b-c\left)\frac{\sqrt{3}}{2}\right|$
$=\sqrt{\left[\right(a-c)+\frac{b-c}{2}{]}^2+[(b-c)\frac{\sqrt{3}}{2}{]}^2}$
Since, $a,b,c$ are integers, not all simultaneously equal.
Let $b=c=k$ and $a=k+1$ then
$|a+b\omega +c{\omega }^2|=\sqrt{[0+1{]}^2+0^2}$
$=1^2+0$
$=1$
Hence, the minimum value of given expression is $1$