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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

$\sqrt{3}$

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B

$\frac{1}{2}$

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C

$1$

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D

$0$

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Solution

## 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|$$\therefore z=\left|a+b\omega +c{\omega }^{2}\right|\phantom{\rule{0ex}{0ex}}⇒{z}^{2}={\left|a+b\omega +c{\omega }^{2}\right|}^{2}\mathbf{\left[}\mathbf{\because }{\mathbf{\omega }}^{\mathbf{2}}\mathbf{+}\mathbf{\omega }\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{\right]}\phantom{\rule{0ex}{0ex}}=\left(a+b\omega +c{\omega }^{2}\right)\left(a+b\overline{\omega }+c{\overline{\omega }}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}+{b}^{2}+{c}^{2}-ac-bc-ab\right)\phantom{\rule{0ex}{0ex}}⇒2{z}^{2}=\left\{{\left(a-b\right)}^{2}+{\left(b-c\right)}^{2}+{\left(c-a\right)}^{2}\right\}$Since, $a,b,c$ are all integer not simultaneously equalIf $a=b$ 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 $\therefore 2{z}^{2}=\left\{{\left(a-b\right)}^{2}+{\left(b-c\right)}^{2}+{\left(c-a\right)}^{2}\right\}\phantom{\rule{0ex}{0ex}}⇒2{z}^{2}\ge \left\{0+1+1\right\}\phantom{\rule{0ex}{0ex}}⇒{z}^{2}\ge 1$Hence, option C is correct.

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