Minimum value of a complex expression involving cube root of a unity

Question

The minimum value of $|a+bw+cw^2|$, where a, b and c are all not equal integers and $w(\ne$1) is a cube root of unity is $$(a) \sqrt{3} \\(b) 1/2 \\(c) 1 \\(d) 0 $$

My attempt: The value $0$ is not possible as it is mentioned a, b and c are all not equal integers. Again, if we assume b = c = 1 then $|a+w+w^2|=|a-1|$ Minimum value of the above expression for integral values of a is $|2-1| = 1$ So option (a) is also ruled out. But in this way I was not able to decide on $1/2.$

I want the right method to solve this question. Any help will be appreciated.

Answer 1

Let $$f(a,b,c) = \left|a+b\omega+c\omega^2\right|\Rightarrow f^2(a,b,c) = \left|a+b\omega+c\omega^2\right|^2 $$

$$=\left(a+b\omega+c\omega^2\right)\cdot\left(a+b\omega^2+c\omega\right) $$ using the fact that: $$\left|z\right|^2=z\cdot \bar{z}$$

Hence, we have: $$f^2(a,b,c)=a^2+ab\omega+ac\omega^2+ab\omega^2+b^2\omega^3+bc\omega^4+ac\omega+bc\omega^2+c^2\omega^3$$ that simplifies to: $$f(a,b,c) = \sqrt{a^2+b^2+c^2-ab-bc-ca} =\sqrt{\frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]} $$

This is minimum when $a=b$ and $\left(b-c\right) ^2 =\left(c-a\right) ^2 =1\implies$ The minimum value is $1$.

Answer 2

$a+b\omega+c\omega^2=a+b\omega-c(1+\omega)=(a-c)+(b-c)\omega$

So, $a+b\omega+c\omega^2=p+q\omega$ for some integers not both equal to $0$.

$|p+q\omega|^2=(p+q\omega)(p+q\bar\omega)=p^2+q^2+pq(\omega+\bar\omega)=p^2-pq+q^2$, which is a positve integer. It's least value is $1$.