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There seems to be this conception that inequalities can't be applied to complex numbers. However, I don't think this is true. It is true that $<$ and $>$ cannot be used between any 2 complex numbers, but we can invent new inequality signs. First, let's focus on $<$. If $a < b$, then $b - a > 0$, and thus, $b - a$ is in the set of positive real numbers. Conversely, $a - b < 0$, and thus, $a - b$ is in the set of negative real numbers. Now, we all know that the argument of a complex number is the angle between the line that that number falls on and the positive real line. We also know that the argument of a positive real number is just $0$. Thus, if $a < b$, $\arg(b - a) = 0$.

Now we can invent a set of new inequality signs. Let us define $<_q$ such that $-p \leq_q p$. For any 2 complex numbers $a$ and $b$, $a <_q b$ implies that $\arg(b - a) = q$. We can see that $<_0$ has the same meaning as $<$. It's important to note that $<$ can be used between non-real numbers, as long as the imaginary parts are the same. If we real numbers $x$, $y$, and $z$ where $x < y$, then $x + zi < y + zi$, since $(y + zi) - (x + zi) = y - x > 0$. Another way to think of this as the if $a <_q b$, then to get from $a$ to $b$, you have to travel parallel to the line whose argument is $q$.

So what do you think? Is this a good way to define inequalities between complex numbers?

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  • $\begingroup$ Why not just say that $\arg a = \arg b = q$ and $|a|<|b|$? Maybe if you are writing up some particular proof of something and you will need to say things like this dozens of times, it might be worth while to define a notation $\stackrel q<$ just for that proof. But to teach everyone to use this notation all the time seems like too much. $\endgroup$ – David K Apr 7 '17 at 20:28
  • $\begingroup$ As it has been discussed here and in some other questions, there is no (total) ordering of the complex numbers which makes it into an ordered field. However, by Zermello's theorem, it can be totally ordered. $\endgroup$ – Pantelis Sopasakis Apr 7 '17 at 20:32
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As you pointed out, the problem isn't about defining inequalities. The problem is about compatibility with natural algebraic operations.

In your case, the main problem is that the order relation isn't total. Meaning that, given any number $a,b\in \mathbf C$, we don't know if there is $a<b$ or $a>b$ or $a=b$. Possibly none of them, and that is a big problem. You can hardly make many computations if you don't have a total order.

However, you will probably be interested by the use of "distances" and "norms" over spaces others than $\mathbf R$. They don't allow to set inequalities between complex numbers, but it allows to establish arguments close enough to what we can do with real numbers (such as triangle inequality : $|a+b| \leq |a| + |b|$).

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So you want to define the size of a complex number as the angle it makes when it is written in polar coordinates?

Consider this:

For any two positive real numbers, we write $a>b$ if $a$ is further from zero than $b$. Visualizing a number line shows this to be fairly obvious.

For a complex number $z$, we use $|z|$ to denote its modulus, or equivalently, distance from zero. We can compare the moduli of two complex numbers, like comparing magnitudes of vectors. It's obvious that $|z|>|w||$ means that $z$ is further from zero than $w$, regardless of the angle it makes with the positive real axis.

I'm trying to argue that there is another perfectly valid way of comparing complex numbers. But the argument is facetious because if there is more than one way to compare, then any comparison is invalid. I don't think I'm doing that good of a job, IMO.

The bottom line is we can compare real numbers because they have one-dimensional ordering (a.k.a. a number line). But complex numbers are two-dimensional, and so they can't be compared in any usual (algebraic) way.

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Of course you could define whatever operations you want, and name it as "$<$", but most of the time, we'd like to see is this an interesting definition? Is it meaningful/compatible? Does it have any nice properties? Will it be helpful for the development of the theories based upon it.

Here is one example of a good definition: Fibonacci Multiplication is an interesting definition as it has nice property (both commutative and associative)

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