Does the complex field satisfy Archimedean Property?

Question

I have two questions regarding the Archimedean Property:

1. I understand that for all $x \in \mathbb{C}$, there exists a positive $n \in \mathbb{N}$ such that $\vert nx \vert >1$ (Please correct me if I'm wrong). So does this mean that the complex number field satisfies the Archimedean property?

2. This is regarding ordered fields. Can it be stated that, WLOG, all ordered fields have the Archimedean property if they do not contain infinitesimal (or infinite) elements, e.g. the Rationals?

Edited: Thanks.

Answer 1

By restricting for $z \in \mathbb C; z \ne 0$ that $|z|$ is such that there is an $n \in \mathbb N$ so that $n|z| > 1$ then you are only considering $|z| \in \mathbb R$. So this has absolutely nothing to do with complex numbers.

To have archimedian property have an sense it must be applied to an ordered set which $\mathbb C$ or $\mathbb R^n$ are not. But we can apply it to the norms of these spaces as the norms are all non-negative real numbers. But this is not considering anything other than $\mathbb R$.

"Can it be stated that, WLOG, all ordered fields have the Archimedean property"

As $\mathbb Q$ is a subfield of every ordered field and the rationals have the archemedian principal. [if $0< q < w < r$ and $q,r\in \mathbb Q$ but $w\not \in \mathbb Q$, then $n*q < n*w < n*r$ so if $1 < n*q$ then $1 < n*w$.

Answer 2

It makes no sense to ask if a field satisfies the Archimedian property unless it is an ordered field. And $\mathbb C$ cannot became an ordered field (basically because $i^2=-1$).