How would you define $>$ and $<$ on $\mathbb{C}$?

It's not at all clear that one of $\mathrm i$ or $-\mathrm i$ is bigger than the other! What makes you think so?

That's a rhetorical question. I know that you think so because of the sign. But there's a big difference between imaginary and real numbers when it comes to signs: $-1$ behaves in a way that is distinctly different from $1$. For instance, if we square $1$, we get back the same number, but if we square $-1$, we obtain a different number. So if I give you a mystery number that is either $1$ or $-1$, you can find out which one it is by asking me what its square is. If you get back the same number it's $1$, otherwise it's $-1$. Such things don't work to differentiate $\mathrm i$ from $-\mathrm i$. If I give you a mystery number that is either $\mathrm i$ or $-\mathrm i$, it will always remain a mystery which one it is, no matter how much you know about its behavior. For instance, if you square the mystery number, you're guaranteed to get $-1$, no matter which of the two it is. If you take the mystery number to the power of $3$, you'll get minus the mystery number, no matter which number it is. Specifically, there is no algebraic way (that is, using only multiplication and addition) to tell the difference. For this reason, there is no algebraic way to tell which of the two should be larger. And when ordering fields, we always want to be able to do so algebraically, since algebra is the only thing we can natively do with fields.

You could define inequalities for complex numbers. For example, we could make the following definition:

We say that $z$ is 'greater than' $w$ if $|z|>|w|$.

The only question that remains is whether this definition is helpful. And the answer to that is: not really. The real numbers are closed under multiplication and addition; that is, if $a$ and $b$ are positive, then $a \cdot b$ is positive, and $a + b$ is positive. It turns out that these two properties are crucially important for defining inequalities in a meaningful way. And neither of these properties hold for complex numbers, meaning that writing $z>w$ just doesn't make that much sense.

To explain why closure under addition and multiplication is relevant, we must first define what we actually mean by $a>b$. One common way to do this is to take the notion of a positive number for granted. Let's denote the collection of positive numbers as $\mathbf{P}$. One of the most important properties of $\mathbf{P}$ is the trichotomy law:

For every number $x$, one and only one of the following holds:

• $x=0$.
• $x$ is in the collection $\mathbf{P}$.
• $-x$ is in the collection $\mathbf{P}$

While this may seem obvious, it is necessary for the following definitions:

• $a>b$ if $a-b$ is in $\mathbf{P}$
• $a<b$ if $b>a$
• $a \geq b$ if $a>b$ or $a=b$
• $a \leq b$ if $b \geq a$

If it were possible for $x$ to be in the collection $\mathbf{P}$, and $-x$ also be in the collection, then we could have $a>b$ and $b>a$ ! Hence, the trichotomy law is necessary if we are to maintain our sanity. Another important property is closure under multiplication and addition, as mentioned earlier. If the real numbers weren't closed under multiplication, this would also cause headaches. For example, we wouldn't be able to say that if $a>0$, and $b>0$, then $ab>0$.

Fortunately, the real numbers do contain a collection $\mathbf{P}$ with all these properties, but the complex numbers do not. This can even be proven fairly easily. Suppose that the complex numbers did satisfy the trichotomy law and closure under multiplication and addition. Let $x$ be an arbitrary complex number. If $x$ is positive, then $xx = x^2$ would also be positive. If $x$ is negative, then $-x$ would be positive, and so $(-x)(-x) = (-x)^2=x^2$ would be positive. It follows that for all non-zero $x$, $x^2$ is positive. In particular, if $x=i$, then $x^2 = -1$ would be positive. But $1=1^2$ would also be positive. This violates the trichotomy law: $1$ and $-1$ can't both be positive.

Hence, the idea of positivity is incoherent for complex numbers, and so too is the idea of inequalities. The best we can do is say that two complex numbers are either equal, or unequal.

You can define $<$ (nearly) anyway you like.

If wanted two I could define $x < y$ if when you spell them out in english words $x$ occurs lower in alphabetical order than $y$.

so $8 < 18 <805 < 85 < 82< 11< 10 < 3 < 0 < 0.5 < 0.7 < 0.6 < 0.1 < 0.2$ and so on

Because eight < eighte < eighth < eightyf < eightyt < el < te < thr < zero < zero point f < zero point se < zero point si < zero point o < zero point t.

Kinda dumb though.

And we could define an order of compairing $a+bi$ to $c + di$ as $a+bi < c+di$ if $a < c$ or if $a=c$ and $b < d$. But $a + bi = c+di$ if $a=c$ and $b=d$. And $a +bi > c+di$ if $a > c$ or if $a=c$ and $b > d$.

That's called the lexigraphical order and it looks okay by it's actually dumb too.

The thing is if we want to define an order we want it to obey in certain ways.

One thing we want is we want if $a < b$ then $a+c < b+c$ for all $c$

The lexigraphical order does do this but what you, incorrectly, claim is the order on the reals does not.

You claim, incorrectly, $-5 > -2$ because $|-5| > |-2|$ and $-5, -2$ both point in the same direction. If this were true we would like $-5 + 10 > -2 + 10$ and that would mean $5 > 8$ which it isn't.

We also want that if $a < b$ and $m > 0$ then $am < bm$.

This fails on the lexigraphical order. $i = 0 + i > 0 + 0i = 0$. so $i*i$ should be greater than $0*i = 0$. but $i*i=-1 < 0$.

And it is because of this requirement we can not have a reasonable order on the complex. That just isn't possible.

Consider: Is $0 < 1$ or is $0 > 1$?

If $0 > 1$ then $0 + (-1) > 1 +(-1)$ and $-1 > 0$. Therefor $(-1)*(-1) > (-1)*0$ so $1 > 0$. But that contradicts what we assumed. So we can't have $0 > 1$ so we must have $0 < 1$.

And therefor $0 + (-1) < 1 + (-1)$ and $- 1 < 0$.

[Note, this means also that if $5 > 2$ the $5-5 > 2-5$ and $0 > -3$ and $0 + (-2) > -3+(-2)$ and so $-2 > -5$. this is why your claim that $-5 > -2$ was wrong. If they point in the negative direction the one with a bigger magnitude must be less than the one with the smaller magnitude.]

Okay, but what about $i$. Is $i > 0$. If so then $i*i > 0*i$ and $-1 > 0$. But we just showed we had to have $-1 < 0$.

Okay, so $i < 0$. But that means $i + (-i) < 0 + (-i)$ so $0 < -i$. So $0*(-i) < (-i)(-i)$. So $0 < (-i)^2 = (-1)^2(i)^2 = 1*(-1) = -1$. So $0 < -1$.... but we just said....

So it's impossible.

If we want $a < b \implies a+c < b+c$ and we also want $a<b; m > 1 \implies am < bm$ then it's just not going to be possible to compare two complex numbers. It just can not be done.