Baby Rudin Chapter 1 Exercise 9: Complex Field ordering Problem: Suppose $z=a+bi$, $w=c+di$. Define $z<w$ if $a<c$, and also if $a=c$ but $b<d$. Prove that this turns the set of all complex numbers into an ordered set. Does this ordered set has the least-upper-bound property?
In my attempt, I've been using Rudin's definition of an ordered field as a proof structure: 
Definition 1.17: An ordered field is a field $F$ which is also an ordered set, such that:
$(1)$ $x+y<x+z$ if $x, y, z \in F$ and $y<z$ 
$(2)$ $xy>0$ if $x\in F, y \in F, x> 0,$ and $y>0$. 
There are some propositions about ordered fields Rudin provides as well which I can prove, but I'm stuck on a part of proving $(2)$.
My attempt on $(2)$: Let $x =a+bi$, $y=c+di$. 
Since $0$ = $(0,0)$ or $0 + 0i$, by the problem's defined ordering, $x>0$ if $a > 0$ OR $a = 0$ and $b>0$. And $y>0$ if the same properties hold for $c, d$. $xy = (ac-bd) + (ad+bc)i$. There are four cases to consider if both $x >0$ and $y>0$:
(1) $a >0, c>0$. 
What if $ac-bd<0$? $ac > 0$ but if $bd > ac$, then $xy$ is not greater than $0 + 0i$.
(2) $a > 0, c = 0$ but $d > 0$.  
$ac = 0, bc = 0$, so $xy = -bd + (ad)i$. If $-bd < 0$, then $xy$ is not greater than $0 + 0i$. 
And so forth. Would appreciate any clarification on this.
For the least-upper-bound property, does it suffice to state than for any non-empty subset $E = \{(c, d)\ :\ c, d \in \mathbb{R}\}$ of the set of complex numbers, we observe that all complex numbers are an ordered pair of real numbers and the real field has the least-upper-bound property so we can always find a complex number $(a, b)$, $a, b \in \mathbb{R}$ such that $a$ and $b$ are supremums of the respective sets $\{c\}$ and $\{d\}$ as defined for subset $E$?
 A: The terminology Rudin chooses here is confusing: an ordered field is not just a field with any order imposed on it (for example, you can put a lexicographic order on the complex numbers). 
The content of the definition is that the order must be compatible in some sense with the field structure, i.e. the field multiplication and addition.
See problem 8. (p. 22) of chapter 1 of Rudin's Principles of Mathematical Analysis: 

Prove that no order can be defined on the complex field which turns it into an ordered field. Hint: -1 is a square.

The real numbers have a natural total order which is compatible with their field structure. The complex numbers do not; this is because $i^2 = -1$, so we get a contradiction if we assign $i$ to be positive ($i >0$) or negative ($i<0$) -- since we want a total order, every number that is not equal to $0$ has to be strictly larger than or strictly smaller than $0$. Again, there is no way to do this which is compatible with the field operations on $\mathbb{C}$.
You will see this elsewhere in mathematics; for example, a topological vector space is not just a vector space which has a topology defined on it; the topology needs to be compatible with the vector space operations.
A: The exercise asks you to show that you get an ordered set. It is not an ordered field; for instance, in the order you gave you have $0<i$, but $i^2=-1<0$, so $(2)$ is not satisfied. 
