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Mathematics Question on Complex Numbers and Quadratic Equations

The minimum value of ∣a+bω+cω2∣| a + b\omega + c \omega^2| , where a,b a, b and cc are all not equal integers and ω (≠1)\omega \, (\ne 1) is a cube root of unity, is

A

3\sqrt 3

B

12\frac{1}{2}

C

1

D

0

Answer

1

Explanation

Solution

Let z=∣a+bω+cω2∣|a+b\omega+c\omega^2|
∣a+bω+cω2∣2=(a2+b2+c2−ab−bc−ca)|a+b\omega+c\omega^2|^2=(a^2+b^2+c^2-ab-bc-ca)
or z2=12(a−b)2+(b−c)2+(c−a)2                 ...(i)z^2=\frac{1}{2} \\{(a-b)^2+(b-c)^2+(c-a)^2\\} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i)
Since, a, b, care all integers but not all simultaneously
equal.
⇒\Rightarrow If a = b then a≠\nec and b ≠\ne c
Because difference of integers = integer
⇒  (b−c)2≥ 1\Rightarrow \, \, (b-c)^2 \ge \, 1 1 {as minimum difference of two consecutive
integers is (±\pm 1)} also (c - a)2≥^2 \ge 1
and we have taken a = b ⇒  (a−b)2=0\Rightarrow \, \, (a-b)^2=0
From E (i), z2≥12(0+1+1)z^2 \ge \frac{1}{2}(0+1+1)
⇒              z2≥1\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, z^2 \ge 1
Hence, minimum value of |z | is 1