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Let $z=a+b\omega + c\omega^2$

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

$$z=a-c+\omega (b-c)$$

Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$

How should I proceed?

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  • $\begingroup$ Good start. Now you have two variables instead of three : $a,b,c$ not equal translates to $x\neq 0,y\neq 0$ where $x=a-c,y=b-c$, and now you have to minimize $|x+y\omega|$ $\endgroup$ May 9, 2020 at 7:13
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    $\begingroup$ Do you mean not all equal, or no two are equal? $\endgroup$ May 9, 2020 at 7:27
  • $\begingroup$ @Displayname this is how it’s written in the question $\endgroup$
    – Aditya
    May 9, 2020 at 8:45

4 Answers 4

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We want to minimize $|x+y\omega|^2=\bigg(x-\frac{y}{2}\bigg)^2+\frac{3}{4}y^2$ where $x$ and $y$ are nonzero integers.

If $y$ is even, we can take $x=\frac{y}{2}$ and hence $|x+y\omega|^2 \geq \frac{3}{4}y^2 \geq 3$.

If $y$ is odd, we can take $x=\frac{y+1}{2}$ and hence $|x+y\omega|^2 \geq \frac{1}{4}+\frac{3}{4}y^2 \geq 1$.

Conclusion : the minimum is $1$. It is reached when $y=\pm 1,x=y$.

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    $\begingroup$ $x=y$ implies $a=b,$ which isn't allowed. We need clarification from the asker if he meant not all equal or no two are equal. $\endgroup$ May 9, 2020 at 7:26
  • $\begingroup$ @Displayname Yes, thanks for the correction $\endgroup$ May 9, 2020 at 7:29
  • $\begingroup$ What does y being even or odd has to do with its relation with x? $\endgroup$
    – Aditya
    May 9, 2020 at 8:49
  • $\begingroup$ @Aditya If $y$ is even, $\frac{y}{2}$ is an integer but $\frac{y\pm 1}{2}$ isn't. If $y$ is odd, $\frac{y\pm 1}{2}$ is an integer but $\frac{y}{2}$ isn't. $\endgroup$ May 9, 2020 at 9:49
  • $\begingroup$ We want $(x-\frac{y}{2})^2$ to be the smallest possible, so we take $x$ to the nearest of $\frac{y}{2}$ $\endgroup$ May 9, 2020 at 9:56
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Following Ewan's comment, note that the span of $\{1,\omega\}$ over $\mathbb{Z}$ is a lattice in the complex plane formed by parallelograms. Drawing the lattice, we see that the closest point to the origin not on the skewed axes and not on the line $z=(1+\omega)t$ is $1+2\omega$ and $|1+2\omega|=\sqrt{3}.$ We also have $|1-\omega|=\sqrt{3}.$

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First of all, my solution is not elegant at all, but I love to think about this problem following way. Although this solution is not that brilliant, I hope someone will find this one useful. Please let me know if you have any questions.

I want to use the concept of elementary dot/inner product of vectors in $\mathbb{R^3}.$ I want to translate the problem following way:

We have three vectors $\vec{P}, \vec{Q},$ and $\vec{R}$ with $ |\vec{P}| = a, |\vec{Q}|=b, $ and $|\vec{R}|=c$. Observe that the angles between any two of the vectors are $120^{\circ}.$ Let $\vec{F}$ be the resultant of the three vectors $\vec{P}, \vec{Q},$ and $\vec{R}$ ie $\vec{F} = \vec{P} +\vec{Q}+\vec{R}.$ Therefore, $\vec{F}.\vec{F} = (\vec{P} +\vec{Q}+\vec{R}). (\vec{P} +\vec{Q}+\vec{R})$ where "$.$" represnents the dot/inner product of vectors in $\mathbb{R^3}.$

Then we have $F^2= P^2 +Q^2+R^2-PQ-QR-RP.$ Equivalently,

$|z|^2 = a^2 +b^2+c^2 -ab-bc-ca= \frac{1}{2}( (a-b)^2+ (b-c)^2 +(c-a)^2).$ Observe that the minimum of $a\sim b$ and $b\sim c$ could be $1.$ This forces $(c-a)^2$ to be $4.$

That means the minimum of |z|^2 is $\frac{1}{2}6=3.$ Hence, the minimum of $|z|$ is $\sqrt{3.}$ QED.

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Note that: $$|a+b\omega+c{\omega}^2|^2=(a+b\omega+c{\omega}^2)(a+b{\omega}^2+c\omega)$$

This is because $\bar{\omega}={\omega}^2$, and $\bar{{\omega}^2}=\omega$. If you multiply out the terms, you'll get $a^2+b^2+c^2-ab-bc-ca$, which is equal to $\frac {(a-b)^2+(b-c)^2+(c-a)^2}{2}$. Since $a,b,c$ are not all equal, two or all three of these square terms cannot be $0$. Hence for minimum let any one of them be $0$ and let the remaining variable differ by $1$. You will get final answer $1$.

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