Prove that field of complex numbers cannot be equipped with an order relation [duplicate]

Please guide me in this problem. I am confused about whether its asking that having the relation $z>0$ does not satisfy the order axioms.

Any help would be really appreciated.

Thanks!

What they are asking is to show that no relation $<$ can exist that complies with the order axioms, i.e.:

1. Only one of $a < b$, $a = b$, or $a > b$ is true
2. If $a < b$ and $b < c$, then $a < c$
3. If $a < b$ and $c < d$, then $a + c < b + d$
4. If $0 < a < b$ and $0 < c < d$, then $a c < b d$

In this case, if we take $i > 0$ we get $i^2 = -1 < 0$, contradicting (4). So by (1) it must be $i < 0$. But $i^4 = 1 > 0$, again contradicting (4). So no relation $<$ can exist on $\mathbb{C}$ which complies with (1) to (4).

try z = i the square root of -1 for a simple contradiction.