Question

Let $\omega \ne 1$ be a cube root of unity. Then the minimum of the set $\left\{\right|a+b\omega +c{\omega }^2{|}^2$ ; Where,$a,b,c$ distinct non-zero integers $\}$ equals.

Answer 1

Finding the minimum value of given set:

Given, set $\left\{\right|a+b\omega +c{\omega }^2{|}^2$; where $a,b,c$ are distinct non-zero integers$\}$

We know that,

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

Now $a,b$ and $c$ are non zero distinct integers. Therefore, for minimum value, we have $a=1,b=2$ and $c=3$

Therefore the minimum of the given set is,

$$\begin{gathered} \frac{1}{2}\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]=\frac{1}{2}\left[{\left(1-2\right)}^2+{\left(2-3\right)}^2+{\left(3-1\right)}^2\right] \\ =\frac{1}{2}\left[1+1+4\right] \\ =\frac{1}{2}\left[6\right] \\ =3 \end{gathered}$$

Therefore, the minimum of a set is equal to $3$