**Puzzle 71**
I think the problem comes down to whether there is a mapping from
$$ \mathbb {C} \rightarrow \mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R} $$
Clearly \\( \mathbb {R} \\) forms a nice poset.
One nice poset is the total order.

I know how to do this with...
$$ \mathbb {Q} \rightarrow \mathbb {Z} \times \mathbb {Z} \rightarrow \mathbb {Z} $$

Here is an idea, interleave the digits of imaginary and real parts of complex number to form a real number. There are as many interleavings as there are bases with is \\(\mathbb{N}\\).

$$ (25.647\dots, 0.023\dots) \rightarrow 2050.604273\dots $$

Another set of ways would be to perform \\(\le\\) in two stages. First look at the \\(\mathfrak{Re}\\) as @MatthewDoty suggested and if that is not strictly less then, and only then, compare \\(\mathfrak{Img}\\).
This approach can be extended to @KeithEPeterson approach, first evaluating the distance from the origin and then comparing the angle, effectively making a spiral.

These ideas involve producing a total order over \\( \mathbb{C} \\) but there are still lots of posets that are not total orders.
The approaches that produce a preorder really only have a problem with the \\( = \\) bit of \\( \le \\).
If we do not require every element to be comparable, not a total order, then we are free to adapt @MatthewDoty approach by allowing \\( = \\) only if the elements are identical. e.g. \\( (2.3 + 1.1i) \le (3.7 - 10.1i) \\) but there is no relation between \\( (2.3 + 1.1i) \text{ and } (2.3 - 10.1i) \\).

Do these approaches work?