Baby Rudin Chapter 1 Exercise 9: Complex Field ordering

The terminology Rudin chooses here is confusing: an ordered field is not just a field with any order imposed on it (for example, you can put a lexicographic order on the complex numbers).

The content of the definition is that the order must be compatible in some sense with the field structure, i.e. the field multiplication and addition.

See problem 8. (p. 22) of chapter 1 of Rudin's Principles of Mathematical Analysis:

Prove that no order can be defined on the complex field which turns it into an ordered field. Hint: -1 is a square.

The real numbers have a natural total order which is compatible with their field structure. The complex numbers do not; this is because $i^2 = -1$, so we get a contradiction if we assign $i$ to be positive ($i >0$) or negative ($i<0$) -- since we want a total order, every number that is not equal to $0$ has to be strictly larger than or strictly smaller than $0$. Again, there is no way to do this which is compatible with the field operations on $\mathbb{C}$.

You will see this elsewhere in mathematics; for example, a topological vector space is not just a vector space which has a topology defined on it; the topology needs to be compatible with the vector space operations.

The exercise asks you to show that you get an ordered set. It is not an ordered field; for instance, in the order you gave you have $0<i$, but $i^2=-1<0$, so $(2)$ is not satisfied.