JEE Advanced 2019 Paper 1Question 1
Let $\omega\neq1$ be a cube root of unity. Then the minimum of the set $$\{|a+b\omega+c\omega^2|^2:\ a,b,c\ \text{distinct non-zero integers}\}$$ equals ____
Complex Numbers
Answer$3$
Solution
$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2-ab-bc-ca=\frac12[(a-b)^2+(b-c)^2+(c-a)^2]$. For distinct integers the minimum is $\frac12(1+1+4)=3$, attained at $a,b,c=1,2,3$.