Example of the norm not acting as a euclidean function in $\mathbb{Z}[\sqrt{-3}]$

Question

It is clear that $\mathbb{Z}[\sqrt{-3}]$ is not a UFD and so cannot be a euclidean domain. However, after some consideration of elements of small norm I have not managed to find the norm failing to act in a euclidean manner. That is, I am looking for an example $a,b \in \mathbb{Z}[\sqrt{-3}] $ such that for all $q,r \in \mathbb{Z}[\sqrt{-3}]$, with $a=bq+r$, have that $N(r)\geq N(b)$ with $N$ the norm.

Answer 1

$R=\mathbb{Z}[\sqrt{-3}]$ is "barely" non-Euclidean (with respect to the usual norm $\sqrt{z\bar{z}}$). Open unit disks centered at elements of $R$ cover $\mathbb{Q}[\sqrt{-3}]$ except for $R+\frac{1+\sqrt{-3}}{2}$. So take any $a$, $b$ with $a/b\in R+\frac{1+\sqrt{-3}}{2}$ to get an example.