Positive and negative complex numbers? Can there be such a thing as positive and negative complex numbers? Why or why not? 
What about positive or negative imaginary numbers? 
It seems very tempting to say $+5i$ is a positive number while $-2i$ is a negative number. On an Argand diagram (complex plane) $+5i$ would be represented by a point above the horizontal axis while $-2i$ is a point below the horizontal axis.
 A: Complex plane does not have a special direction. That's why in my opinion. Real line on the other hand, does. It's similar to the fact that time has a direction, and space doesn't.
I would also argue against "positive" imaginary numbers. Since we defined $i$ as the solution of equation $x^2 =  -1$ note that $-i$ is also a solution. Thus, we can't really distinguish between $i$ and $-i$. 
A: You can never make complexes a linear ordered field, since any ordered field is formally real field
A: A complex number can be treated as a kind of multivector existing in a 2-d geometric algebra (GA).
Seen this way, the answer to your question is that you can indeed negate complex numbers, much in the same way that multivectors (or just plain old vectors) can be negated.
Changing the sign of the imaginary component alone is equivalent to 'complex conjugation', or in GA parlance, 'reversion'. If the complex number is of unit length, reversion is identical to inversion.
A: [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.
A: 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 $i^2 < 0$.
A: 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 $-1+1=0$
A: 'Positive' and 'Negative' are defined only on the real number line, which is part of the system of complex numbers.  So it makes sense to say, for example $1 -100i$ is positive and $-1 + 100i$ is negative, based upon their real number values.
Although arbitrary, there is also some sense of a positive and negative imaginary numbers. While it is impossible to tell the two apart, especially if the values are supposed to be tied somehow to the real world, in the system of complex numbers, they are different: $[(5+i)x +(1+3i)]^2$ has a different solution than $[(5-i)x+(1+3i)]^2$ does. ($24x^2+{\color{red}4}x-8+(10x^2+32x+6)i$ vs. $24x^2+{\color{red}{16}}x-8-(10x^2-28x-6)i$.) That does not make $6i$ a positive number, however; it is a positive imaginary number. It has no real value and as such is only 'positive' in an completely fictional sense.
A: I would like to add some insight given to me by someone called “Somiaj”.

The major issue is that real numbers are the largest Archimedian field (complete and linearly ordered which means they obeys certain properties like if $a<b$ then $a+c < b+c$). You can of course put an infinite number of these linear orders on the complex numbers and they would be ordered but they would not preserve this order when you start adding/subtracting things, much less multiplication, which is a big part of what we like about the reals.

A: $\left|-1\right| = 1,$ a rotation of $\pm\pi$
$|1| = 1,$ a rotation of $0$
$\left|-i\right| = 1,$ a rotation of $-\pi/2$
$|i| = 1,$ a rotation of $\pi/2$
ambiguity in rotational sign line 1 and no rotation in line 2 for real
no ambiguity in lines 3,4; and equal magnitude, opposite rotation between 3,4 for imag
very different, non-correlated behavior of |x| on real and imaginary, would seem to throw crowbar into the whole attempt to answer affirmative.
A: We may partially order the complex numbers in accord with their real parts so that 
   a+bi < c+di iff a < c. 
Thus the line of imaginary numbers divides positive complex numbers from negative complex numbers.  However, if b and d are not equal, a+bi and c+di are incomparable.  Thus it is no longer true that every positive number is greater than every negative number.  Most positive-negative pairs are incomparable.  This partial ordering has some neat features.  For example, in the Riemann sphere representation of the complex plane, a point at infinity is added to the imaginary line, transforming it into a great circle.  This circle thus divides the Riemann sphere into positive and negative hemispheres.  The imaginary numbers, 0, and infinity, which lie between the hemispheres, are all “neutral”—that is, neither positive nor negative—but all other complex numbers are either positive or negative. 
