Question

If $a,b,c$ are integers, no two of them being equal and $\omega$ is complex cube root of unity, then minimum value of $|a+b\omega +c{\omega }^2|$ is

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

Answer 1

The correct option is D $\sqrt{3}$

Let

$z=a+b\omega +c{\omega }^2$

Since

$\overline{\omega }={\omega }^2$,

and vice versa, we have

$\overline{z}=a+b{\omega }^2+c\omega$

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

$=(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )$

$=(a^2+b^2+c^2-ab-bc-ca)$

$=\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$

Hence,

$|a+b\omega +c{\omega }^2|$$=\frac{1}{\sqrt{2}}\sqrt{\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]}$

The minimum occurs when t the integers are consecutive, i.e if they are of the form

$a=b-1$

$c=b+1$

Thus, the minimum value:

$|a+b\omega +c{\omega }^2|$$=\frac{1}{\sqrt{2}}\sqrt{[1+1+4]}$

$=\sqrt{3}$