Complex Numbers cannot be Ordered Compatibly with Ring Structure

Theorem
There exists no ordering relation on the field of complex numbers $$\left({\mathbb{C}, +, \times}\right)$$ compatible with the structure of $$\left({\mathbb{C}, +, \times}\right)$$.

Proof
Suppose the opposite, and that the relation $$\preceq$$ is such that:


 * $$z \ne 0 \Longrightarrow 0 \prec z \lor 0 \prec -z$$, but not both;
 * $$0 \prec z_1, z_2 \Longrightarrow 0 \prec z_1 z_2, 0 \prec z_1 + z_2$$.

Then since $$\mathbb \imath \ne 0$$, it follows that $$0 \prec \imath$$ or $$0 \prec -\imath$$.

Suppose $$0 \prec \imath$$.

Then $$0 \prec \imath \cdot \imath = -1$$.

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

Then $$0 \prec \left({-\imath}\right) \cdot \left({-\imath}\right) = -1$$.

So whichever, we have $$0 \prec -1$$.

But at the same time, $$0 \prec \left({-1}\right)^2 = 1$$.

So we have both $$0 \prec -1$$ and $$0 \prec 1$$ which contradicts the hypothesis that $$\prec$$ is an compatible with the structure of $$\left({\mathbb{C}, +, \times}\right)$$.

Hence there can be no such ordering.