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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.

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up vote 4 down vote accepted

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 $i^2 < 0$.

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[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.

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When you said "it's the other way around" are you implying $i$ and $-i$ are the same number? – hb20007 May 9 '14 at 17:45
@hb20007 : No. I'm saying the choice of which one is called $i$ and which is called $-i$ does not matter. – Michael Hardy May 9 '14 at 17:46
I think you should expand on what exactly "nothing at all would change" means and how you justify it. Something to do with there being an automorphism interchanging them, I suppose. – Jack M May 9 '14 at 23:42
@JackM : One could mention the word "automorphism", but in fact I figured this out long before I ever encountered that word. – Michael Hardy May 10 '14 at 15:18
In fact, engineers have taken this to the extreme of saying that $j=-i$ is the real square root of -1, and changed all their formulas accordingly. (Well, sort of. They care mostly about time dependence, so they prefer the form $e^{j\omega t}$, whereas spatially-minded physicists prefer $e^{i(kz-\omega t)}$ for plane waves. And they feel $i$ is an appropriate symbol for current, so they shifted over to $j$. They assure me it makes sense.) Whatever the reasons, there are large stretches of physics vs engineering formula mismatches which are magically fixed by setting $j=-i$. – E.P. May 5 '15 at 20:20

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 $-1+1=0$

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In other words: squares are positive, and everything is a square. – jmc May 10 '14 at 11:18

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 $-i$.

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You can never make complexes a linear ordered field, since any ordered field is formally real field

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You probably meant "formally real", not "formally $\mathbb R"$ (which would mean "formally the field of real numbers"). – Andreas Blass May 9 '14 at 19:35
@AndreasBlass yes :) I edited it – Math137 May 9 '14 at 22:36

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.

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'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.

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It would be better to say the imaginary part of $6i$ is positive, because the imaginary part of a complex number is (technically) the real number that is the coefficient of $i$. – hardmath May 10 '14 at 5:11

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."

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|-1| = 1, a rotation of +/-pi

|1| = 1, a rotation of 0

|-i| = 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.

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