Question - Find the minimum value of the following expression where a b c are distinct non zero integers and $\omega \neq 1$ is a cube root of unity . $$|a+b\omega+c\omega^2|^2.$$

I used triangle inequality for this problem. $||a+b\omega|-|c\omega^2||\leq $$|a+b\omega+c\omega^2|^2.$

But $$|a+b\omega|\leq|a|+|b| \cdots(|\omega|=1)$$

Therefore $$||a|+|b|-|c||\leq |a+b\omega+c\omega^2|.$$

But minimum value of $$||a|+|b|-|c||=0$$

Therefore $$|a+b\omega+c\omega^2|^2_{min}=0?.$$

But answer given is 3. Have i applied triangle inequality correctly for 3 conplex numbers?

  • $\begingroup$ I don't understand your reasoning. You have shown that the thing which you are trying to minimize is $≥$ something that might be small. But that doesn't prove anything. We already knew that your expression was $≥0$. $\endgroup$
    – lulu
    Commented Nov 5, 2023 at 15:39
  • $\begingroup$ @lulu okay. Can you give ideas on how to approach this problem? $\endgroup$
    – Aleph
    Commented Nov 5, 2023 at 15:54
  • $\begingroup$ $\omega^2 + \omega + 1 = 0$ although they say distinct. I would just experiment with small coefficients and $\omega = \frac{-1 + i \sqrt 3}{2} $ $\endgroup$
    – Will Jagy
    Commented Nov 5, 2023 at 15:57
  • 1
    $\begingroup$ Hint: Consider a hexagonal grid where the axis are $1, \omega, \omega^2$. $\endgroup$
    – Calvin Lin
    Commented Nov 5, 2023 at 16:01
  • $\begingroup$ @CalvinLin wow that is very interesting . I have never thought of it like that $\endgroup$
    – Aleph
    Commented Nov 5, 2023 at 16:09

2 Answers 2


Posting my method as demanded by @Calvin Lin .

$|a+b\omega+c\omega^2|^2= (a +b\omega+c\omega^2)(\overline{a+b\omega+c\omega^2)}=(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$

Now ,

$(a +b\omega+c\omega^2)(\overline{a+b\omega+c\omega^2)}=(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^2+b^2+c^2-ab-bc-ca.$

So now we have to minimize the above expression -


$a,b,c \in\mathbb{Z^+} \implies$ WLOG put $a=1 , b=2 , c=3. \implies$

$$\boxed{|a +b\omega+c\omega^2|^2_{min}=3}$$

  • 1
    $\begingroup$ +1 looks good. $\quad$ To clarify, the "demand" for you to post your solution is because MSE closes questions for which OP doesn't show their work. $\endgroup$
    – Calvin Lin
    Commented Nov 7, 2023 at 14:52
  • Using a hexagonal grid, let's consider where the lattice point $(a, b, c) = a+ b \omega + c \omega^2$ lands up in the complex plane.
    • The hexagonal grid is helpful for thinking about this problem, but not crucial.
  • Since $ 1 + \omega + \omega^2 = 0 $, we know that $(a, b, c) = (a-k, b-k, c-k)$.
    • In this problem, the condition that these values are non-zero is a red herring, since we could just add the same value to each coordinate. We thus remove that restriction from the problem. (Alternatively, we solve this problem for a larger set of points and show that the minimum doesn't change when we restrict to the original set of points.)
    • This coordinate shift doesn't change the condition that the integers are pairwise distinct, which is crucial to the problem.
  • Let's figure out what the minimum distance of these distinct-coordinate lattice points are.
    • The closest lattice point is at distance 0, namely the origin $(0,0,0)$. However, it cannot be achieved by distinct coordinates .
    • The next closest lattice points are at distance 1, and are $(1, 0, 0), (-1, 0, 0)$ and their cyclic permutations. Again, these cannot be achieved by distinct coordinates.
    • The next closest lattice points are at distance $\sqrt{3}$, and are of the form $(0, 1, -1)$, $(0,-1, 1)$ (and their cyclic permutations). These can be achieved by distinct coordinates.
  • Thus, the minimum of the expression is $3$. Equality is achieved by $(0+k, 1+k, -1+k), (0+k, -1 + k, 1 + k)$ (and their cyclic permutations).
  • $\begingroup$ +1 Great solution , thanks. $\endgroup$
    – Aleph
    Commented Nov 7, 2023 at 15:49

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