Question

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

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

Answer 1

The correct option is C

$1$

Explanation for correct option :

Given expression is $\left|a+b\omega +c{\omega }^2\right|$where $a,b$and $c$ are all not equal integers and $\omega \ne 1$

We know that cube root of unity is $\omega$

That is $x^3-1=\left(x-1\right)\left(x^2+x+1\right)$

Then ${\omega }^3=1$ and the root of the equation $x^2+x+1=0$ is $\omega ,{\omega }^2$

From the equation $x^2+x+1=0$ we know, $1+\omega +{\omega }^2=0$.

Let $z=\left|a+b\omega +c{\omega }^2\right|$

$$\begin{gathered} \therefore z=\left|a+b\omega +c{\omega }^2\right| \\ \Rightarrow z^2={\left|a+b\omega +c{\omega }^2\right|}^2\mathbf{[}\mathbf{\because }{\mathbf{\omega }}^{\mathbf{2}}\mathbf{+}\mathbf{\omega }\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{]} \\ =(a+b\omega +c{\omega }^2)(a+b\overline{\omega }+c{\overline{\omega }}^2) \\ =\left(a^2+b^2+c^2-ac-bc-ab\right) \\ \Rightarrow 2z^2=\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\} \end{gathered}$$

Since, $a,b,c$ are all integer not simultaneously equal

If $a=b$ then $a\ne c$ and $b\ne c$

Because difference of the integer is integer

$\therefore {\left(b-c\right)}^2\ge 1$

As minimum difference of two consecutive integers is $+1$

Similarly ${\left(c-a\right)}^2\ge 1$

Take ${\left(a-b\right)}^2=0$

We have

$$\begin{gathered} \therefore 2z^2=\left\{{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right\} \\ \Rightarrow 2z^2\ge \left\{0+1+1\right\} \\ \Rightarrow z^2\ge 1 \end{gathered}$$

Hence, option C is correct.