Question

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

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

Answer 1

Answer: (C)

Let $z=|a+b\omega +c{\omega }^2|$
$|a+b\omega +c{\omega }^2{|}^2=(a^2+b^2+c^2-ab-bc-ca)$
or $z^2=\frac{1}{2}\{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}...(i)$
Since, $a,b,c$ are all integers but not all simultaneously equal.
$\Rightarrow$ If $a=b$ then $a\ne c$ and $b\ne c$
Because difference of integers = integer
$\Rightarrow (b-c{)}^2\ge 1$ {as minimum difference of two consecutive integers is ($\pm 1$)} also $(c-a{)}^2\ge 1$
and we have taken $a=b\Rightarrow (a-b{)}^2=0$
From Eq. (i), $z^2\ge \frac{1}{2}(0+1+1)$
$\Rightarrow z^2\ge 1$
Hence, minimum value of$|z|$ is $1$