Can we determine if a complex number is greater than another?

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.

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.

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.