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. Then
$a+b\omega +c{\omega }^2=\left|\begin{array}{c}a+b\left(\frac{-1+i\sqrt{3}}{2}\right)+c\left(\frac{-1-i\sqrt{3}}{2}\right)\end{array}\right|=\left|\begin{array}{c}a+b\left(\frac{2a-b-c}{2}\right)+i\left(\frac{b\sqrt{3}-c\sqrt{3}}{2}\right)\end{array}\right|=\frac{1}{2}\sqrt{(2a-b-c{)}^2+3(b-c{)}^2}=\frac{1}{2}\sqrt{(4a^2+b^2+c^2-4ab+2bc-4ac+3b^2+3c^2-6bc)}=\sqrt{a^2+b^2+c^2-ab-bc-ca}=\sqrt{\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]}$
R.H.S. will be minmum when a=b=c, but we cannot take a=b=c as per the question. Hence, the minimum value is obtained when any two are zero and third is a minimum magnitude integer, i.e., 1. Thus, b=c=0, a=1 gives us the minimum value of 1.