Question

Consider that $a,b,c$ are distinct integers and $\omega \ne 1$ is a cube root of unity, then minimum value of \[x=\left| a+b\omega +c{{\omega }^{2}} \right|+\left| a+b{{\omega }^{2}}+c\omega \right|\] is

Answer 1

Hint: Initially, we can assume \[{{z}_{1}}=a+b\omega +c{{\omega }^{2}}\] and then we can find its conjugate i.e. \[\overline{{{z}_{1}}}\] . To proceed ahead we should know the following properties:
1) \[{{\left| {{z}_{1}} \right|}^{2}}={{z}_{1}}.\overline{{{z}_{1}}}\]
2) \[\left| {{z}_{1}} \right|=\left| \overline{{{z}_{1}}} \right|\]
3) \[\overline{\omega }={{\omega }^{2}}\]
4) \[\overline{{{\omega }^{2}}}=\omega \]
5) \[{{\omega }^{3}}=1\]
6) \[1+\omega +{{\omega }^{2}}=0\]

Complete step by step answer:
Let us assume
\[{{z}_{1}}=a+b\omega +c{{\omega }^{2}}\text{ }.............\text{(1)}\]
So, the conjugate of \[{{z}_{1}}\] will be given by
\[\begin{align}
  & \overline{{{z}_{1}}}=\overline{a+b\omega +c{{\omega }^{2}}} \\
 & \Rightarrow \overline{{{z}_{1}}}=a+b\overline{\omega }+c\overline{{{\omega }^{2}}} \\
\end{align}\]
Now putting \[\overline{\omega }={{\omega }^{2}}\] and \[\overline{{{\omega }^{2}}}=\omega \] in the above equation, we get
\[\Rightarrow \overline{{{z}_{1}}}=a+b{{\omega }^{2}}+c\omega \text{ }.............\text{(2)}\]
Now we know that \[{{\left| {{z}_{1}} \right|}^{2}}={{z}_{1}}.\overline{{{z}_{1}}}\]
Putting the value of \[{{z}_{1}}\] and \[\overline{{{z}_{1}}}\] in this equation, we get
\[\begin{align}
  & {{\left| {{z}_{1}} \right|}^{2}}=\left( a+b\omega +c{{\omega }^{2}} \right)\left( a+b{{\omega }^{2}}+c\omega \right) \\
 & \Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{a}^{2}}+ab{{\omega }^{2}}+ca\omega +ab\omega +{{b}^{2}}{{\omega }^{3}}+bc{{\omega }^{2}}+ca{{\omega }^{2}}+bc{{\omega }^{4}}+{{c}^{2}}{{\omega }^{3}} \\
 & \Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{a}^{2}}+{{b}^{2}}{{\omega }^{3}}+{{c}^{2}}{{\omega }^{3}}+ab\left( \omega +{{\omega }^{2}} \right)+bc\left( {{\omega }^{4}}+{{\omega }^{2}} \right)+ca\left( \omega +{{\omega }^{2}} \right) \\
\end{align}\]
We can put \[{{\omega }^{4}}=\omega .{{\omega }^{3}}\] in the above equation
\[\Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{a}^{2}}+{{b}^{2}}{{\omega }^{3}}+{{c}^{2}}{{\omega }^{3}}+ab\left( \omega +{{\omega }^{2}} \right)+bc\left( \omega .{{\omega }^{3}}+{{\omega }^{2}} \right)+ca\left( \omega +{{\omega }^{2}} \right)\]
We can put \[{{\omega }^{3}}=1\] in the above equation
\[\Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab\left( \omega +{{\omega }^{2}} \right)+bc\left( \omega +{{\omega }^{2}} \right)+ca\left( \omega +{{\omega }^{2}} \right)\]
We know that \[1+\omega +{{\omega }^{2}}=0\] , so we can put \[\omega +{{\omega }^{2}}=-1\] in the above equation
\[\Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca\]
Multiplying L.H.S and R.H.S with 2 in the above equation we get
\[\begin{align}
  & \Rightarrow 2{{\left| {{z}_{1}} \right|}^{2}}=2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca \\
 & \Rightarrow 2{{\left| {{z}_{1}} \right|}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-2ab \right)+\left( {{b}^{2}}+{{c}^{2}}-2bc \right)+\left( {{c}^{2}}+{{a}^{2}}-2ca \right) \\
\end{align}\]
Applying the identity \[{{x}^{2}}+{{y}^{2}}-2xy={{\left( x-y \right)}^{2}}\] in the above equation, we get
\[\Rightarrow 2{{\left| {{z}_{1}} \right|}^{2}}={{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}}\]
Dividing L.H.S and R.H.S by 2 in the above equation we get
\[{{\left| {{z}_{1}} \right|}^{2}}=\dfrac{\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]}{2}\]
Taking the square root of the above equation, we get
\[\left| {{z}_{1}} \right|=\sqrt{\dfrac{\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]}{2}}\text{ }................\text{(3)}\]
Now, According to the question
\[x=\left| a+b\omega +c{{\omega }^{2}} \right|+\left| a+b{{\omega }^{2}}+c\omega \right|\]
Combining the above equation with equation (1) and equation (2), we get
\[x=\left| {{z}_{1}} \right|+\left| \overline{{{z}_{1}}} \right|\]
Putting \[\left| \overline{{{z}_{1}}} \right|=\left| {{z}_{1}} \right|\] in the above equation, we get
\[\Rightarrow x=2\left| {{z}_{1}} \right|\]
Putting the value of \[\left| {{z}_{1}} \right|\] in the above equation from equation (3), we get
\[\Rightarrow x=2\sqrt{\dfrac{\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right]}{2}}\text{ }............\text{(4)}\]
We have to find the minimum value of \[x\] , so the difference between \[a\] and \[b\] , \[b\] and \[c\] , \[c\] and \[a\] should be minimum and as all are integers so the minimum difference can be 1. Since, $a,b,c$ are distinct integers, therefore, difference between all the three pairs might not be same.
So, let us assume that
 $a=n$
$b=n+1$
$a=n+2$
Putting the above values in equation (4), we get
\[\begin{align}
  & \Rightarrow x=2\sqrt{\dfrac{\left[ {{\left( n-\left( n+1 \right) \right)}^{2}}+{{\left( n+1-\left( n+2 \right) \right)}^{2}}+{{\left( n+2-n \right)}^{2}} \right]}{2}} \\
 & \Rightarrow x=2\sqrt{\dfrac{\left[ {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}} \right]}{2}} \\
 & \Rightarrow x=2\sqrt{\dfrac{\left[ 1+1+4 \right]}{2}} \\
 & \Rightarrow x=2\sqrt{\dfrac{6}{2}} \\
 & \Rightarrow x=2\sqrt{3} \\
\end{align}\]

So, the minimum value of \[x\] has come out to be \[2\sqrt{3}\]

Note: The thing that you should notice in this question is that the expression given in the question contains conjugate pairs and their modulus will give equal value. There is a slight possibility that you might make a mistake while calculating the minimum value of \[x\] by using the fact that the minimum value of the square of the difference of two integer is zero and to avoid that mistake you have to consider that all the integers are distinct and thus their difference can never be zero.