Comparing complex numbers

We can define a partial order on $\Bbb C$ by $z_1\prec z_2$ if and only if $|z_1|<|z_2|$.

We can define a total order on $\Bbb C$ in various ways--I'll give you a few if you're interested.

We cannot give an order that is compatible with the operations on $\Bbb C$ so that $\Bbb C$ is an ordered field. If we could, would $i$ be positive or negative?

Basically, it depends on how you want to define "bigger/smaller" in this instance. Give us more detail, and we can better answer your question.

You can't define a total order $\le$ so that $\mathbb{C}$ is an ordered field:

$(1)\space\forall a, b, c \in \mathbb{C}, a ≤ b \Rightarrow a + c ≤ b + c$

$(2)\space\forall a, b \in \mathbb{C}, 0 ≤ a \land 0 ≤ b \Rightarrow 0 ≤ a b$

Either $0 \le 1$ or $1 \le 0$ in which case $0\le-1$. Let $\varepsilon \in\{-1,1\}$ so that $0 \le \varepsilon$

Either $0\le i$ or $i \le 0$ in which case $0\le-i$. Let $\delta \in \{-i,i\}$ so that $0\le\delta$

Now you can just derive something absurd:

You have $0\le \varepsilon$ and $0\le \delta$

So by $(2)$, $0\le\varepsilon\delta$

By $(2)$ again, $0\le\varepsilon\delta\delta = \varepsilon(-1) = -\varepsilon$

So $\varepsilon \le 0$

Now we have $0\le\varepsilon$ and $\varepsilon\le 0$ so $\varepsilon=0$

But we have $\varepsilon\in\{-1,1\}$

Absurd.