There is an exercise "Prove that no order can be defined in the complex field that turns it into a field." in baby Rudin's book. And I can get a contradiction if $i>0$ or $i<0$. However, I wonder if we can say $i=0$, even more, all complex numbers all equal. I think this order actually satisfies all the condition:
(i) If $x\in S$ and $y\in S$ then one and only one of the statements $$x<y, x=y, y<x$$ is true.
(ii) If $x, y, s\in S$, if $x<y$ and $y<z$, then $x<z$.
And
(i) $x+y<x+z$ if $x,y,z\in F$ and $y<z$,
(ii) $xy>0$ if $x\in F$, $y\in F$, $x>0$, and $y>0$.
Am I right? Can we say this is an order?
One more question, is singleton an ordered set?
No. That doesn't make sense, because when we define a binary relation $R$ on a set $S$, it is assumed that we are dealing with the standard concept of $=$ on $S$. For instance, we say that $R$ is anti-symmetric if$$x\mathrel Ry\wedge y\mathrel Rx\implies x=y.$$