Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 3

Theorem
Let $\left({\C, +, \times}\right)$ be the field of complex numbers.

There exists no total ordering on $\left({\C, +, \times}\right)$ which is compatible with the structure of $\left({\C, +, \times}\right)$.

Proof
From Complex Numbers form Integral Domain, $\left({\C, +, \times}\right)$ is an integral domain.

suppose that $\left({\C, +, \times}\right)$ can be ordered.

Thus, by definition, it possesses a positivity property $P$.

Then from Positivity Property induces Total Ordering, let $\le$ be the total ordering induced by $P$.

From Unity of Ordered Integral Domain is Positive:
 * $1$ is positive.

Thus by positivity, axiom $3$:
 * $-1$ is not positive.

Consider the element $i \in \C$.

By definition of positivity, axiom $3$, either:
 * $i$ is positive

or:
 * $-i$ is positive.

Suppose $i$ is positive.

Then by Square of Element of Ordered Integral Domain is Positive:
 * $i^2 = -1$ is positive.

Similarly, suppose $-i$ is positive.

Then by Square of Element of Ordered Integral Domain is Positive:
 * $\left({-i}\right)^2 = -1$ is positive.

In both cases we have that $-1$ is positive.

But it has already been established that $-1$ is not positive.

Hence, by Proof by Contradiction, there can be no such ordering.