Let $\omega \neq 1$ be a cube root of unity. Then the minimum of the set $$\left\{\left|\mathrm{a}+\mathrm{b} \omega+\mathrm{c} \omega^{2}\right|^{2}: \mathrm{a}, \mathrm{~b}, \mathrm{c} \text { distinct non-zero integers }\right\}$$ equals ____

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Complex Numbers

Let

ω \neq = 1

be a cube root of unity. Then the minimum of the set

{a + b ω + c ω^{2}^{2} : a, b, c distinct non-zero integers}

equals ____

Answer: 3

For a cube root of unity

ω

, we have

\overline{ω} = ω^{2}

and

1 + ω + ω^{2} = 0

a + bω + c ω^{2}^{2} = (a + bω + c ω^{2}) (\overline{a + bω + c ω^{2}})

= (a + bω + c ω^{2}) (a + b ω^{2} + c ω)

= a^{2} + b^{2} + c^{2} + ab (ω + ω^{2}) + b c (ω^{3} + ω^{3}) + c a (ω^{2} + ω)

= a^{2} + b^{2} + c^{2} - ab - b c - c a

= \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]

For distinct non-zero integers, the minimum occurs when the differences are as small as possible. Taking

a = 1, b = 2, c = 3

\frac{1}{2} [1^{2} + 1^{2} + 2^{2}] = \frac{1}{2} (1 + 1 + 4) = 3

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer