This post did not quite seem to answer my question. If we have
$\quad a+bi\in\mathbb{C}\quad$ can ordering be done the way of Pythagorean triples $\quad A^2+B^2=C^2?\quad$
There are infinite triples for each of
$\quad
(C-B),\space
(C-A),\space\text{and}\space
(B\pm A)\quad$
but these may be further ordered as sets of sets
$
(C-B)=(2n-1)^2\in\big\{1,3,5,\cdots\big\}
$
$
(C=A)=2k^2\in\big\{2,8,18,\cdots\big\}
$
\begin{align*}
&(B\pm A)=P:\\
&P=p_1,\cdots ,p_n),\space p_k\equiv \pm 1 \text{ (mod }8) \in\big\{1,7,17, \cdots\big\}
\end{align*}
For each of $(A\text{ or } B\space\text{ or } \space C),\space$
there are $2^{n-1}\space$
primitive triples where $\space n\space $ is the number of unique prime factors of $\space(A\text{ or } B\space\text{ or } \space C)\space $ and these may be of vastly different sizes. Perhaps these may be ordered by the pairs of natural numbers $\space(n,k)\space$ that generate them in this formula
\begin{align*}
A=(2n-1)^2+ & 2(2n-1)k \tag{a} \\
B= \qquad\qquad\quad & 2(2n-1)k+ ]] 2k^2 \tag{b} \\
C=(2n-1)^2+ & 2(2n-1)k+ 2k^2 \tag{c}
\end{align*}
where $\space n\space $ is a set number and $\space k\space$ is the ordinal triple within each set.
For example, $C=65\space$ represents triples
$\space f(4,1)=(63,16,65)\quad f(5,4)=(33,56,65).\space$ where $\space(63,16,65)\space$ is the first member of $Set_4$ and $\space(33,56,65)\space$ is the fourth member of $Set_5.$
Can $\space (4+1i),\space (5+4i)\space$ be treated as natural ordering of these complex numbers or are their other considerations such as the size of the resulting squares or products to consider?
Best Answer
There are an infinite number of ways to order the complex numbers as a set.
There is absolutely no way to order the complex numbers as a field.
I didn't bother to read your explanation of the ordering of pythogorean triples but it seems similar to "lexigraphic ordering".
For two complex number as numbers $w,z$ we can define the lexigraphic order as:
(In other words: If $w = a+bi; z = c+di; a,b,c,d \in \mathbb R$ to compare $w$ to $z$ compare $a$ to $c$. If $a = c$ then compare $b$ to $d$.)
That satisfies everything we need to have an ordered set. To be an order on a set we need:
Given complex $w$ and $z$ then exactly one of the following is true: Either $w \prec z$ or $z\prec w$ or $w = z$.
if $w\prec z$ and $z\prec \alpha$ then it always follows that $w \prec \alpha$.
It's easy to see that those are both always true. So $\prec$ is an order on $\mathbb C$ as a set.
......
But it does not satisfy an order on a field.
To be an order on a field we need to be able to do math on these numbers and have the resulting inequalities be preserved. We need:
This is actually true.
and
This fails. This fails big time!
Let $w = 4 - 6i$ and $z = 5-3i$ and $a = i$.
Then $4 - 6i \prec 5-3i$ and $0\prec i$ but $(4-6i)i = 6 + 4i \not\prec 3 + 5i = (5-3i)i$.
.....
Now having an order as a field is impossible.
Why. Consider any $w$ in a field. We either have $w > 0$ or $w = 0$ or $w< 0$.
If $w > 0$ then $w^2 = w\times w > 0$. If $w=0$ then $w^2 = 0$. And if $w < 0$ then $0 =w +(-w) < 0 + (-w) =-w$ and so $w^2 = (-w)\cdot (-w) > 0$.
So we have $w^2 \ge 0$ and we never have $w^2 < 0$.
But in the complex we have $1^2 = 1$ so we would have to have $1^2 = 1 > 0$. Then we'd have to have $0 < 1$ so $-1 =0+(-1) < 1 +(-1) = 0$ so $-1 < 0$.
But we also have And we have $i^2 = -1$ so we would HAVE to have $i^2 = -1 > 0$. So we have BOTH $-1 < 0$ AND $-1 > 0$. That's utterly impossible.
So the complex numbers can not be ordered as a field.