Extended Triangle Inequality to $3$ Complex Numbers

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?

• 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 $≥0$.
– lulu
Commented Nov 5, 2023 at 15:39
• @lulu okay. Can you give ideas on how to approach this problem? Commented Nov 5, 2023 at 15:54
• $\omega^2 + \omega + 1 = 0$ although they say distinct. I would just experiment with small coefficients and $\omega = \frac{-1 + i \sqrt 3}{2}$ Commented Nov 5, 2023 at 15:57
• Hint: Consider a hexagonal grid where the axis are $1, \omega, \omega^2$. Commented Nov 5, 2023 at 16:01
• @CalvinLin wow that is very interesting . I have never thought of it like that Commented Nov 5, 2023 at 16:09

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 -

$$a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$$

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

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

• +1 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. 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).
• +1 Great solution , thanks. Commented Nov 7, 2023 at 15:49