Completion of a number field at a complex embedding

Question

Sorry if this question has been asked before.

Let $K$ be a number field of degree $n>1$ and $\sigma:K\hookrightarrow \mathbb C$ a complex (non real) embedding of $K$ in $\mathbb C$ giving the absolute value $|x|_\sigma=|\sigma(x)|$ where $|-|$ is the usual absolute value on $\mathbb C$.

My question is, why is the completion of $K$ with respect to $|-|_\sigma$ equal to $\mathbb C$?

Do I use the fact that $\mathbb Q \subset K \subset \mathbb C$ and the completion of $\mathbb Q$ is $\mathbb R$?

I'll be very grateful for a proof or references.

Answer 1

A number field $K$ of degree $n=r+2s$ has $r$ real embeddings and $2s$ complex ones. Since $\sigma$ is not real, i.e., $s>0$, the completion of $K$ cannot be $\mathbb{R}$. Hence it must be $\mathbb{C}$, since it can only be $\mathbb{R}$ or $\mathbb{C}$ with respect to an Archimedian metric (or a $p$-adic field for a Non-Archimedean metric).