Can we determine if a complex number is greater than another? Is it possible to determine if one complex number is greater than another? Or as the question implies is there an "order" to complex numbers (like 1 is before 2 in the real numbers)?
I thought that would could simply use the modulus to determine if one complex numbers is greater than another, though I believe this can't be the only way used (what if 2 complex numbers have the same modulus  and are quite different). So I thought, if the point is in the uppermost right quadrant of the complex plane, then both real and imaginary parts are positive, so it would be greater than any other complex number in a different quadrant of the complex plane (you might say what about the modulus, but in the reals $1>-2$ even though $|-2|>|1|$). But what if one complex number has a positive real, and negative imaginary and another one has a negative real and positive imaginary? (And for arguments sake they both have the same modulus)
If we can't determine why not? In the real numbers it seems (to me), quit trivial at a basic level to determine if one real is greater than another e.g. $2>1$. What is this property of numbers called? Why doesn't complex numbers exhibit this property (if indeed it doesn't)?
 A: The field $\mathbb{C}$ has only a total order compatible with addition.  It has no total order which is compatible with multiplication, prohibiting it from being an ordered field.  Finally, its usual topology cannot be generated by any total order.
A: If $i>0$, then $i^2>0 \Rightarrow -1>0$ which is a contradiction.
If $i=0$, then $i^2=0 \Rightarrow -1=0$ which is a contradiction.
If $i<0$, then $i^2>0 \Rightarrow -1>0$ which is a contradiction.
Thus $ib$ is neither greater nor equal nor less than $0$.
So any complex number $a+ib$, ($a,b \in \mathbb{R}$)  is neither greater nor equal nor less than $a$.
And the reasoning follows.
A: It is possible to order the complex numbers. For instance, one could define $x_1+iy_1<x_2+iy_2$ if $x_1<x_2$ or if $x_1=x_2$ and $y_1<y_2$.
However, it's impossible to define a total order on the complex numbers in such a way that it becomes an ordered field. This is because in an ordered field the square of any non-zero number is $>0$. Hence we would have $-1=i^2>0$, and adding $1$ to both sides would imply $0>1=1^2$, which is a contradiction.
A: There is no total order on $\mathbb{C}$ compatible with the order on $\mathbb{R}$ and compatible with the algebraic operations. Suppose there was such an order, then either $i>0$ or $i<0$. If $i>0$, then multiplying by $i$ we get that $-1=i^2>0$ which is impossible. If $i<0$, then multiplying by $i$ reverses the inequality, and so we get that $-1=i^2>0$. Both lead to contradictions.
