showing there is an element of $\mathbb{Z}[\beta]$ is in the unit ball of every complex number

Question

Let $\beta=\frac{1+i\sqrt{3}}{2}$, and let $\mathbb{Z}[\beta]=\{a+b\beta:a,b\in\mathbb{Z}\}$. I'm trying to show that for every $x\in \mathbb{C}$ there exists a $y\in\mathbb{Z}[\beta]$ such that $\lvert x-y \rvert<1$, where $\lvert x \rvert$ is the complex norm.

I tried thinking of $\mathbb{Z} [\beta]$ as a lattice on the complex plane and tried to show that every rectangle had diagonals with length less than or equal to 2 but couldn't exactly work out how I can think of $\mathbb{Z} [\beta]$ as a lattice. I would perhaps approach it a different way. I would appreciate any help.

Answer 1

Call your complex number $z$ and write it as $z=r+si\sqrt{3}$, $r$, $s$ real. Pick the nearest integer to $2s$, $b$ say. Consider $z-b(1+i\sqrt3)/2 =r'+s'i\sqrt 3$. Then $|s'|\le 1/4$. Pick the nearest integer to $r'$, $a$ say. Define $\gamma=a+b\beta$. Show that $|z-\gamma|^2\le (1/2)^2 +3(1/4)^2$, etc.

Answer 2

A demonstration based upon $E \upsilon \kappa \lambda \iota \delta \eta \varsigma$:

Consider if you will he points

$A = 0, \; B = 1,\; C = 1 + \beta,\; D = \beta \in \Bbb C; \tag 1$

they form a quadrilateral--a rhombus, actually, $ABCD$ in the complex plane. We note that $\beta = \frac{1}{2}(1 + i \sqrt 3)$ implies that

$\angle BAD = \dfrac{\pi}{3}, \tag 2$

and it follows from the law of cosines that diagonal $BD$ of $ABCD$ has length given by

$\vert BD \vert^2 = \vert AB \vert^2 + \vert AD \vert^2 - 2\vert AB \vert \vert AD \vert \cos \dfrac{\pi}{3} = 1 + 1 - 2 (\dfrac{1}{2}) = 2 - 1 = 1; \tag 3$

thus

$\vert BD \vert = 1; \tag 4$

likewise, since

$\angle ABC = \dfrac{2\pi}{3}, \tag 5$

we have

$\vert AC \vert^2 = \vert AB \vert^2 + \vert BC \vert^2 - 2\vert AB \vert \vert BC \vert \cos \dfrac{2\pi}{3} = 1 + 1 - 2(-\dfrac{1}{2}) = 3, \tag 6$

whence

$\vert AC \vert = \sqrt 3. \tag 7$

Now let $\bigodot A$ be the circle of radius $1$ centered at $A$ and let $\bigodot C$ be that of radius $1$ centered at $C$; since each of these circles is of unit radius, $\bigodot A$ and $\bigodot C$ intersect precisely at points $B$ and $D$, and since $\vert AC \vert = \sqrt 3 < 2$, every point on $ABCD$ or in its interior lies in the interior of one of these two circles, save points $B$ and $D$, which lie on the the circles themselves. Thus if we include $\bigodot B$ and $\bigodot D$, each also of radius $1$, every point on the rhombus $ABCD$ lies in the interior of one of the circles $\bigodot A$, $\bigodot B$, $\bigodot C$, $\bigodot D$. But this is tantamount to saying that each point $X$ in or on $ABCD$ satisfies $\vert XY \vert < 1$ for some $Y \in \{ A, B, C, D \}$, which statement is equivalent to $\vert x - y \vert < 1$ for $x \in \Bbb C$ in or on $ABCD$ and some $y \in \{ 0, 1,\beta, 1 + \beta \} \subset \Bbb C$. Translating the rhombus $ABCD$ and $\bigodot A$, $\bigodot B$, $\bigodot C$, and $\bigodot D$ by $m + n \beta$ for $m, n \in \Bbb Z$ covers the entire complex plane with congruent rhombuses and identical circles, and from this the result follows.