Can the complex numbers be totally ordered, and if so, how is this different to $\mathbb{C}$ being an ordered field?
We can totally order $\mathbb{C}$ can we not, for example by saying that:
$$a>c\implies (a+bi>c+di) \quad\forall a,b,c,d,$$
and by complementing that with:
$$(a=c)\land (b>d)\implies (a+bi>c+di) \quad\forall a,b,c,d$$
But $\mathbb{C}$ is not an ordered field so clearly I am missing something somewhere; presumably in what exactly it means to be an ordered field.
By your ordering, $i > 0$ and $1 > 0$. But if that ordering made $\mathbb{C}$ an ordered field, then
\begin{align*} &i > 0\\[6pt] \implies\; &i^2 > 0\\[6pt] \implies\; &-1 > 0\\[6pt] \implies\; &-1 + 1 > 0 + 1\\[6pt] \implies\; &0 > 1 \end{align*}
contradiction.