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.