Positive and negative complex numbers?

[begin crackpot theory]

Mathematicians have committed a great blunder by thinking that a certain imaginary number is $i$ and another is $-i$, when it's really the other way around!!!!

[end crackpot theory]

If one were to take the position above, and try to rewrite all of mathematics consistently with the theory propounded above, the result would be that nothing at all would change. Which one is called $i$ and which is called $-i$ doesn't matter.

It does in some contexts make sense to pay attention to whether the real part of a complex number is positive. A Dirichlet series $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^s}$ converges if the real part of $s$ is greater than the real part of the abscissa of convergence of the series, and diverges if the real part of $s$ is less than the abscissa of convergence.

You may turn $\mathbb{C}$ into a totally ordered set and then define $a \geq b$ if and only if $a-b \geq 0$. An example of such a total order on $\mathbb{C}$ is the lexicographic order defined on $\mathbb{R}^2$.

The problem is that if you do so, then when you restrict this new order to $\mathbb{R}$ as a subset of $\mathbb{C}$ and you expect it to satisfy the field order axioms, this order will not agree with the usual ordering on $\mathbb{R}$. So, you can't think of this order as an extension of the order we have on $\mathbb{R}$.

In fact it is not possible to define an order on $\mathbb{C}$ that interacts with addition and multiplication the same way that elements of $\mathbb{R}$ do. This is because that by accepting the field order axioms on $\mathbb{R}$, you can prove that $\forall x \in \mathbb{R}: x^2 \geq 0$, while this theorem breaks in $\mathbb{C}$ because $i^2 < 0$.

Suppose that $\Bbb C=P\cup N \cup \{0\}$, this union being disjoint.
I assume that you want that $P+P=P$, $PP= P$, $NN= P$ and $PN=N$.

So $-1=ii$ must live in $P$. And $(-1)(-1)=1$ must also live in $P$. But $-1+1=0$