Show that there doesn't exist a relation $\succ$ between complex numbers such that
(i) For any two complex numbers $z,w$, one and only one of the following is true: $z\succ w,w\succ z,$ or $z=w$
(ii) For all $z_1,z_2,z_3\in\mathbb{C}$ the relation $z_1\succ z_2$ implies $z_1+z_3\succ z_2+z_3$.
(iii) For all $z_1,z_2,z_3\in\mathbb{C}$ with $z_3\succ 0$, then $z_1\succ z_2$ implies $z_1z_3\succ z_2z_3$.
Suppose $i\succ 0$. From (iii) we have $i^2\succ 0$, so $-1\succ 0$, so applying (ii) we get $0\succ 1$. But repeating (iii) on $-1\succ 0$ we get $1\succ 0$, a contradiction. So either $i=0$ or $0\succ i$.
How can I proceed from here?
Best Answer
If we had an order on the complex numbers, then either $i \prec 0$ or $0 \prec i$.
If $0 \prec i$, then $$0i \prec ii \implies 0 \prec -1$$
Then since $0 \prec -1$, we see that $0 \prec (-1)^2 = 1$. Using (iii) we get
$$0 \prec -1 \implies 1 = 0 + 1 \prec -1 + 1 = 0 \implies 1 \prec 0 \prec 1$$
contradicts (i). The case that $i \prec 0$ is similar. Just use (ii) and add $(-i)$ both sides.