Complex Number inqualities Although the inequalities are not defined on complex numbers. But does the inequality $x < 4 + 5i$  be said to possess any solutions ? Where $ i = \sqrt{-1}$.
 A: The impossibility for $\mathbb{C}$ to be an ordered field is a general result of fields theory.

An ordered field $F$ is a field with a total order relation $\le$
  compatible with the field operations. This means that $\forall a,b,c
> \in F$ : $$ a \le b \Rightarrow a+c \le b+c $$ $$ 0\le a \,,\; 0 \le b
> \Rightarrow 0 \le ab $$

We can show that this definition is equivalent to:
There exists a subset $\mathcal{P} \subset F$ such that: 

$\forall a,b \in  \mathcal{P}$ we have  $a+b \in  \mathcal{P}$ and $ab
> \in  \mathcal{P} $  (closure property)

and

$\forall a \in  A$ only one of these relations hold: $ a \in 
> \mathcal{P}$, or $ a=0$ , or $-a \in  \mathcal{P}$  ( trichotomy
  property)

From this two properties we can easely see that : $1 \in \mathcal{P}$ and $\forall a \in F\,,\,a \ne 0  \Rightarrow (-a)^2=a^2 \in \mathcal{P}$, so:
a field can not be an ordered field if it contain an element $i$ such that $i^2=-1$.
Note that this does not means that we can not ghive any order on $\mathbb{C}$.  As an exeple we can define the lexicographical order:
$$
x=a+ib \le y=c+id \iff (a<c)\; \lor\; (a=c \land b\le d)
$$
whith wich we can solve inequalities as $x< 4+5i$, but this order is not compatible with the operations in $\mathbb{C}$.
