Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 1

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
there exists a relation $\preceq$ on $\C$ which is ordering compatible with the ring structure of $\C$.

That is:


 * $(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both
 * $(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$

By Totally Ordered Ring Zero Precedes Element or its Inverse, $(1)$ can be replaced with:


 * $(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both.

As $i \ne 0$, it follows that:
 * $0 \prec i$ or $0 \prec -i$

Suppose $0 \prec i$.

Then:

Otherwise, suppose $0 \prec \left({-i}\right)$.

Then:

Thus by Proof by Cases:
 * $0 \prec -1$

Thus it follows that:

Thus both:
 * $0 \prec -1$

and:
 * $0 \prec 1$

This contradicts hypothesis $(1')$:
 * $(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both

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