I have the following assignment:
consider the map
$$|\cdot|:\mathbb{Z}[i]\longrightarrow \mathbb{N},\qquad |a+ib|:=a^2+b^2$$
1) Prove that $|\alpha|<|\beta|$ iff $|\alpha|\leq |\beta|-1$ and $|\alpha|<1$ iff $\alpha=0$
2) Let $\alpha,\beta\in\mathbb{Z}[i],\beta\neq 0$. Prove that the map $f:\mathbb{Z}[i]\longrightarrow\mathbb{Z}[i], f(\gamma):=\alpha-\gamma\beta$ is the composition of a dilatation by the factor $\sqrt{|\beta|}$, a rotation (angle?) and a translation.
3) Deduce that there exists $\gamma\in\mathbb{Z}[i]$ such that $|f(\gamma)|$ is strictly smaller than $|\beta|$.
$\textbf{Hint:}$ compare the size of a cell of the lattice $f(\mathbb{Z}[i])$ and the size of the set of points whose distance to $0$ is $\leq\sqrt{|\beta|}$.
What i did: point 1) is a trivial consequence of the fact that the norm takes integer non negative values. For point 2), I use complex multiplication of numbers which is: multiply absolute values and add angles. For point 3), i'm actually waiting for a miracle... I suppose i should prove that there exists a cell in $f(\mathbb{Z}[i])$ intersecting the open ball centered at the origin with radius $\sqrt{|\beta|}$, but i have no idea how to write down this. Only thing i noticed is that $f$ acts with a rotation, which does not affect distance from the origin, so that the only changes in $|\gamma|$ come from dilatation and by adding $\alpha$.
Could someone put me on the right direction?
Hints (or not...)
1) This follows at once from the fac that $\;\forall\alpha\in\Bbb Z[i]\;,\;\;|\alpha|\in\Bbb N\,$
2) We have that
$$f=f_\alpha\circ f_\gamma\;,\;\;\text{with}\;\;f_\gamma(x):=\gamma x\;,\;\;f_\alpha(x)=\alpha-x$$
3) The closed unit circle centerd at the origin is compact and thus there're only a finite number of elements of any lattice (say, $\,\Bbb Z[i]\,$ ...?) intersecting it
Now you can dilate (shrink, in scientific talk...:) ) elements as shown in (2) so that their inside that circle...
The basic assumption here is that $\,\beta\neq0\,$ , I presume...