Can an ordered field contain complex numbers? I read a question about ordering of complex numbers, and saw an answer showing that there cannot exist an ordering of the complex numbers because regardless of how $i$ is placed in that order, it would imply that $i^2 = -1$ would be positive.
This proof can obviously be generalized to prove that for any non-zero $a \in Q$ it is impossible for $ai$ to be an element of an ordered field. But there is no obvious generalization of the proof showing that the same hold for all imaginary numbers.
Does there exist a sub-field of the complex numbers containing at least one imaginary number, which can be assigned a consistent ordering?
Or stated in different terms, does there exist a subset of the complex numbers, which is a formally real field and is not a subset of the reals?
 A: This is an old question but since it is unanswered I thought I would provide an interesting example of such a field (technically it is answered: Daniel Fischer's answer suffices as far as I can tell). Another reason I wanted to provide this answer as opposed to the more elegant one given in Daniel Fischer's comment is to highlight the fact that $\mathbb{C}$ contains some pretty strange subfields. 
Let $C(\mathbb{R})$ be the ring of continuous real-valued functions defined on $\mathbb{R}$ and let $I' \subseteq C(\mathbb{R})$ be the ideal consisting of functions with compact support, with $I$ the corresponding maximal ideal. Consider the residue field $C(\mathbb{R})/I$. It is proven here (rather famously) that this is a real-closed field containing a proper subfield isomorphic to $\mathbb{R}$, so there is a unique total order $\leq$ we can define on this field. We now note that $F=C(\mathbb{R})/I$ has cardinality $\mathfrak{c}$:
Proof: It is obvious that $|F| \geq \mathfrak{c}$ since $F$ contains a subfield isomorphic to $\mathbb{R}$. To prove the other inequality, note that $F$ is actually a collection of equivalence classes, i.e. $F= \left\{ [f]: f \in C(\mathbb{R}) \right\}$, for some equivalence relation. By the axiom of choice, there is a function $g:F \rightarrow C(\mathbb{R})$ such that $g([f]) \in [f]$ for each $f$. Such a map is necessarily an injection, so $|F| \leq |C(\mathbb{R})|=\mathfrak{c}$. Schroeder-Bernstein then gives our result. $\Box$
So $F$ has the cardinality of the continuum, and it is also easy to see that $F$ has characteristic $0$. Since any field of characteristic zero and cardinality at most continuum is isomorphic to a subfield of $\mathbb{C}$ (stated here and I've also seen this invoked in other places as well), we have $F \cong K \subseteq \mathbb{C}$ for some field $K$. Give $K$ the order induced by the isomorphism with $F$. Since $F$ contains a proper subfield isomorphic to $\mathbb{R}$ and $\mathbb{R}$ contains no such thing, we know $K$ is not contained in $\mathbb{R}$, so $K$ is a field satisfying your desired properties.
Now, if you are familiar with the paper I linked above by Hewitt, you'll recognize that this will actually work for any non-compact, completely regular topological space $X$ with a maximal ideal $I$ such that $C(X)/I$ has cardinality $\mathfrak{c}$ and has the so-called eta set property. We can find infinitely many such topological spaces and ideals, thus it is tempting to believe that we can actually produce infinitely many non-isomorphic examples of such fields in the way we did above. A word of caution about this: it is completely consistent with ZFC that all fields produced in this manner are isomorphic (see here, it is an implication of the continuum hypothesis). Under CH, the unique (up to isomorphism) field we have constructed is a "hyperreal number field" and happens to be the most common one used in non-standard analysis.
