The ordering on $\mathbb{R}$ is "compatible with the operations."
This means the following: there is a subset $P$ of $\mathbb{R}$, called the "positive number", such that:
- $\mathbb{R}$ obeys a trichotomy law: every $r\in \mathbb{R}$ satisfies one and only one of $r\in P$, $-r\in P$, and $r=0$.
- If $a,b\in P$, then $a+b\in P$ and $ab\in P$.
Then one defines $a\lt b$ if and only if $b-a\in P$, and $a\leq b$ if and only if $a=b$ or $a\lt b$.
With these definitions, we obtain the "usual" properties : if $a\leq b$ then for all $c$ we have $a+c\leq b+c$; if $a\leq b$ and $c\leq d$ then $a+c\leq b+d$; if $a\leq b$ and $c\gt 0$, then $ac\leq bc$. Etc. That is, obtain an "ordered field".
However, there is no way to define an order on $\mathbb{C}$ that is compatible with the operations. That is, there is no subset $P$ of $\mathbb{C}$ that will satisfy properties 1 and 2 above.
To see why, note that if we had a class of "positive complex numbers", then we must have $1$ positive: because either $1\in P$ and we are done, or else $-1\in P$, and then $1=(-1)(-1)\in P$. So, either way, $1\in P$. But because $1\in P$, we must have $-1\notin P$.
Now, if $i\in P$, then $-1=(i)(i)\in P$, which contradicts trichotomy. And if $i\notin P$, then $-i\in P$, so $(-i)(-i)=-1\in P$... again contradicting trichotomy. Thus, there is no class of "positive complex numbers" satisfying 1 and 2 above, and hence no order on $\mathbb{C}$ that is compatible with the operations.
Because of that, when dealing with the algebraic side of $\mathbb{C}$, such as when working with vector spaces over $\mathbb{C}$, we do not put an order on all of $\mathbb{C}$. We still have the order on $\mathbb{R}$, which we can use to compare magnitudes of complex numbers, or real numbers. So when we have a complex number $z\in\mathbb{C}$, and a real number $r$, we only write $z\geq r$ if $z$ is real, and $z\geq r$ as real numbers. In particular, if we write $\langle z,z\rangle\geq 0$, then we are saying that $\langle z,z\rangle$ is real and nonnegative.
(You can define all sorts of orders on $\mathbb{C}$; for example, the lexicographic order on the complex plane gives $a+bi\lt c+di$ if and only if $a\lt c$, or $a=c$ and $b\lt d$; but this order does not play well with addition and multiplication; and no ordering on $\mathbb{C}$ plays well with addition and multiplication, so when you care about the algebraic structure on $\mathbb{C}$, you do not put an order on it except perhaps for ad hoc arguments.)
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