If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal.

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?

• 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 $|x+y\omega|$ May 9, 2020 at 7:13
• Do you mean not all equal, or no two are equal? May 9, 2020 at 7:27
• @Displayname this is how it’s written in the question May 9, 2020 at 8:45

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 $$y=\pm 1,x=y$$.

• $x=y$ implies $a=b,$ which isn't allowed. We need clarification from the asker if he meant not all equal or no two are equal. May 9, 2020 at 7:26
• @Displayname Yes, thanks for the correction May 9, 2020 at 7:29
• What does y being even or odd has to do with its relation with x? May 9, 2020 at 8:49
• @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. May 9, 2020 at 9:49
• We want $(x-\frac{y}{2})^2$ to be the smallest possible, so we take $x$ to the nearest of $\frac{y}{2}$ May 9, 2020 at 9:56

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 $$|1-\omega|=\sqrt{3}.$$

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.

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 $$1$$.