Puzzle 71. I would like to give a try. I guess we have to focus on modulus or real and imaginary parts separately, considering, in a geometrical sense, distances from the origin on the complex plane, and numbers as tuples of reals.
Let \\(z_1=a_1+i b_1\\) and \\(z_2=a_2+i b_2\\) be two complex numbers. Let us consider, for example, \\(|z_1|\le |z_2|\\). Let \\(z_3=a_3+i b_3\\) and \\(z_4=a_2+i b_4\\) be another pair of complex numbers. Let us suppose that they verify \\(|z_3|\le |z_4|\\). Thus, we get \\(|z_1|+|z_3|\le |z_2|+|z_4|\\). The 0 could work both for modulus as well as for real and imaginary parts separately: \\((a_1+0)=a_1=(0+a_1)\\) and \\((b_1+0)=b_1=(0+b_1)\\); but also \\(|z_1|+0=|z_1|=0+|z_1|\\). I would say that \\(|z_1|\le|z_2|\\) and \\(|z_2|\le|z_1|\\) implies \\(|z_1|=|z_2|\\), and \\((a_1)\le(a_2)\\) and \\((a_2)\le(a_1)\\) implies \\(a_1=a_2\\); similarly for the imaginary part. Also, if \\(a_1\le a_2\\) and \\(a_2\le a_3\\), thus \\(a_1\le a_3\\) -- and \\(b_1\le b_2\\), \\(b_2\le b_3\\) implies \\(b_1\le b_3\\). Moreover, \\((a_1+a_2)+a_3=a_1+(a_2+a_3)\\), as well as \\((b_1+b_2)+b_3=b_1+(b_2+b_3)\\). This is also true for the modulus: \\(|z_1|\le |z_2|\\) and \\(|z_2|\le |z_3|\\) implies \\(|z_1|\le |z_3|\\), and \\((|z_1|+ |z_2|)+|z_3|=|z_1|+(|z_2|+|z_3|)\\). Finally, \\(a_1+a_2=a_2+a_1\\), \\(b_1+b_2=b_2+b_1\\), and \\(|z_1|+|z_2|=|z_2|+|z_1|\\).

We might just work with real and imaginary parts separately. We can write \\(z_1\\) and \\(z_2\\) as \\((a_1,b_1)\\) and \\((a_2,b_2)\\), respectively. If, comparing the real part of the first number with the real part of the second number, and the imaginary part of the first number with the imaginary part of the second number, we get \\(a_1 \le a_2\\) and \\(b_1 \le b_2\\), it implies that \\((a_1,b_1)\le(a_2,b_2)\\). This would be graphically equivalent to considering the distance of points in the complex plane from the origin. Let \\(z_3=a_3+i b_3\\) and \\(z_4=a_4+i b_4\\) be other two complex numbers, that we can write as \\((a_3,b_3)\\) and \\((a_4,b_4)\\), respectively. If \\(a_3\le a_4\\) and \\(b_3\le b_4\\), then \\((a_3,b_3)\le(a_4,b_4)\\).
We can compare the first pair of numbers, \\(z_1,z_2\\), with the second pair of numbers, \\(z_3,z_4\\). In particular, if \\((a_1+a_3)\le(a_2+a_4)\\) and \\((b_1+b_3)\le(b_2+b_4)\\), then \\(((a_1+a_3),(b_1+b_3))\le((a_2+a_4),(b_2+b_4))\\).