JEE Challenger
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Minimum Value of Complex Expression with Cube Root of Unity

Suppose ω≠1\omega \neq 1 is a complex cube root of unity. Find the minimum value of the set {∣a+bω+cω2∣2:a,b,c are distinct non-zero integers}\{|a + b\omega + c\omega^2|^2 : a, b, c \text{ are distinct non-zero integers}\}

Official Numerical Answer3

Step-by-Step Solution

To find the minimum value of the expression ∣a+bω+cω2∣2|a + b\omega + c\omega^2|^2 for distinct non-zero integers a,b,ca, b, c, we first simplify the modulus using the properties of the complex cube root of unity ω\omega.

Since ω=−12+i32\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} and ω2=−12−i32\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}, we can expand the squared magnitude as follows: ∣a+bω+cω2∣2=(a+bω+cω2)(a+bω2+cω)|a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) =a2+b2+c2−ab−bc−ca= a^2 + b^2 + c^2 - ab - bc - ca =12[(a−b)2+(b−c)2+(c−a)2]= \frac{1}{2} \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]

Since a,b,ca, b, c are distinct non-zero integers, without loss of generality, assume a<b<ca < b < c. The minimal possible absolute differences between consecutive distinct integers are b−a≥1b - a \ge 1 and c−b≥1c - b \ge 1, which implies c−a≥2c - a \ge 2.

Substituting the minimum possible differences b−a=1b - a = 1, c−b=1c - b = 1, and c−a=2c - a = 2 (which is satisfied by distinct non-zero integers such as a=1,b=2,c=3a = 1, b = 2, c = 3), we obtain: Minimum value=12[12+12+22]=12(1+1+4)=3\text{Minimum value} = \frac{1}{2} \left[ 1^2 + 1^2 + 2^2 \right] = \frac{1}{2} (1 + 1 + 4) = 3

Thus, the minimum value of the set is 33.