Question

That $\omega \ne 1$ be the cube root of unity. Then the minimum of the set ${{\left| a+b\omega +c{{\omega }^{2}} \right|}^{2}}$where a, b and c are distinct non-zero integers is equal to?

Answer 1

First, before proceeding for this, we must assume the given function as z and then we get:
$z=a+b\omega +c{{\omega }^{2}}$. Then, by using the property of the complex number which says that $\overline{\omega }={{\omega }^{2}},\overline{{{\omega }^{2}}}=\omega$, we get the solvable expression. Then, for minimum conditions let us assume the minimum natural numbers as values which are non- zero as a=1, b=2 and c=3, we get the minimum value.

Complete step-by-step answer :
In this question, we are supposed to find the minimum value of ${{\left| a+b\omega +c{{\omega }^{2}} \right|}^{2}}$ where a, b and c are distinct non-zero integers.
So, before proceeding for this, we must assume the given function as z and then we get:
$z=a+b\omega +c{{\omega }^{2}}$
Then, by multiplying the complex number z with its complex conjugate $\overline{z}$,we get:
$z\times \overline{z}=\left( a+b\omega +c{{\omega }^{2}} \right)\left( \overline{a+b\omega +c{{\omega }^{2}}} \right)$
Now, by using the property of the complex number which says that $\overline{\omega }={{\omega }^{2}},\overline{{{\omega }^{2}}}=\omega$.
So, by using the property stated above, we get the expression as:
$\left( a+b\omega +c{{\omega }^{2}} \right)\left( a+b{{\omega }^{2}}+c\omega \right)$
Then, by solving the above expression, we get:
\begin{aligned} & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \\\ & \Rightarrow \dfrac{1}{2}\left\\{ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right\\} \\\ \end{aligned}
Now, we are given in the question that a, b and c have distinct non-zero values.
So, for minimum conditions let us assume the minimum natural numbers as values which are non- zero as:
a=1, b=2 and c=3 which gives the minimum value.
So, by substituting the values assumed above in the obtained function to get the minimum value as:
\begin{aligned} & \dfrac{1}{2}\left\\{ {{\left( 1-2 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}+{{\left( 3-1 \right)}^{2}} \right\\} \\\ & \Rightarrow \dfrac{1}{2}\left\\{ {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}} \right\\} \\\ & \Rightarrow \dfrac{1}{2}\left\\{ 1+1+4 \right\\} \\\ & \Rightarrow \dfrac{1}{2}\left\\{ 6 \right\\} \\\ & \Rightarrow 3 \\\ \end{aligned}
So, we get the minimum value as 3 for the given question.
Hence, 3 is the minimum value for the set ${{\left| a+b\omega +c{{\omega }^{2}} \right|}^{2}}$where $\omega \ne 1$ be the cube root of unity.

Note : Now, to solve these types of the questions we need to know some of the basic conversions beforehand so that we can easily proceed in these types of questions. Then, some of the basic conversions are:
$\left( {{a}^{2}}+{{b}^{2}}-2ab \right)={{\left( a-b \right)}^{2}}$
Moreover, we must be careful while dealing with these types of questions as complex numbers behave differently as compared to normal numbers.