QJEEAdv19Macc7c: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Macc7c Solution: Let be a cube root of unity. Then the minimum of the set…

Given, $\omega \ne 1$ be a cube root of unity, then ${\left| {a + b\omega + c{\omega ^2}} \right|^2}$

= $(a + b\omega + c{\omega ^2})\overline {(a + b\omega + c{\omega ^2})}$ ,

[ $\because$ $z\overline z = |z{|^2}$ ]

= $(a + b\omega + c{\omega ^2})$ ${(a + b\overline \omega + 2c{{\overline \omega }^2})}$

[ $\because$ ${\overline \omega }$ = $\omega$ ² and ${{{\overline \omega }^2}}$ = $\omega$ ]

= ${a^2} + {b^2} + {c^2} + ab({\omega ^2} + \omega ) + bc({\omega ^2} + {\omega ^4}) + ac(\omega + {\omega ^2})$

[as ${\omega ^3} = 1$ ]

$= {a^2} + {b^2} + {c^2} + ab( - 1) + bc( - 1) + ac( - 1)$

[as $\omega + {\omega ^2} = - 1,\,{\omega ^4} = \omega$ ]


$= {a^2} + {b^2} + {c^2} - ab - bc - ca$

$= {1 \over 2}\{ {(a - b)^2} + {(b - c)^2} + {(c - a)^2}\}$

$\because$ a, b and c are distinct non-zero integers. For minimum value a= 1, b = 2 and c = 3

$\therefore$ $|a + b\omega + c{\omega ^2}|_{\min }^2 = {1 \over 2}\{ {1^2} + {1^2} + {2^2}\} = {6 \over 2} = 3.00$