Let omega neq 1 be a cube root of unity. Then | vedclass

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(B) We know that $|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(\overline{a + b\omega + c\omega^2})$.Since $\overline{\omega} = \omega^2$ an

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$3$

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(B) We know that $|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(\overline{a + b\omega + c\omega^2})$.Since $\overline{\omega} = \omega^2$ and $\overline{\omega^2} = \omega$,this becomes $(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.Expanding this,we get $a^2 + ab\omega^2 + ac\omega + ab\omega + b^2\omega^3 + bc\omega^2 + ac\omega^2 + bc\omega^4 + c^2\omega^3$.Using $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,this simplifies to $a^2 + b^2 + c^2 - ab - bc - ca$.This expression can be written as $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.For distinct non-zero integers $a, b, c$,the smallest possible values for the differences are $1$ and $2$ (e.g.,$a=1, b=2, c=3$).Substituting these,we get $\frac{1}{2}[(1 - 2)^2 + (2 - 3)^2 + (3 - 1)^2] = \frac{1}{2}[1 + 1 + 4] = \frac{6}{2} = 3$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

Let $\omega \neq 1$ be a cube root of unity. Then the minimum value of the set $\{|a + b\omega + c\omega^2|^2 : a, b, c \text{ are distinct non-zero integers}\}$ is equal to:

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