Let $\omega
eq 1$ be a cube root of unity. Then the minimum of the set $\left\{\left|a+b \omega+c

Question

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

Professor PC Thomas Classes · Answer

(3) : $\left|a+b \omega+c \omega^{2}ight|^{2}$\ $=\left(a+b \omega+c \omega^{2}ight)\left(a+b \omega^{2}+c \omegaight)$\ $=a^{2}+b^{2}+c^{2}-a b-b c-c a$\ $=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}ight]$\ $=\frac{1}{2}(1+1+4)=3$\ Since it will be minimum only if $a, b, c$ are consecutive integers.\