Total ordering on complex numbers 
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?
 A: There is a well-known geometric interpretation of complex multiplication as a scaling followed by a rotation on a vector in the plane.
In brief, if $z_1$ is a complex number represented in the polar form $|z_1|e^{i \theta}$, then multiplying $z_2$ by $z_1$ is equivalent to scaling $z_2$ by $|z_1|$ and rotating it counterclockwise by $\theta$.
Here is a proof using this idea. Suppose $0 \prec 1$.
In the picture links 'red' means '$\succ$' and 'blue' means '$\prec$'.
Multiplying the values on sides by the complex unit vectors $e^{i \theta}$ is equivalent to rotating them by $\theta$. When we do so for $\theta \in [0, 2 \pi]$ 1 becomes the unit circle, while 0 remains the same. Because our ordering is preserved by complex multiplication, everything on the unit circle $\succ$ 0.
On the other hand, adding a real number $x$ to both sides is equivalent to translating the values along the x-axis. When we translate 0 and 1 by all of the real numbers $x$, we find that everything to the 'right' of zero $\succ$ 0, and everything to the 'left' of zero $\prec$ 0.
By inspecting the pictures, we see that complex multiplication forces $-1 \succ 0$, while complex addition forces $-1 \prec 0$. A contradiction!
With a little work this can be made rigorous. The idea also informs a variety of alternative proofs for this theorem. For example, the accepted answer uses rotation by unit vectors and translation along the imaginary axis to provide the contradiction.
A: 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.
A: $i\ne 0$ since $i$ has an inverse but $0$ does not. Now just note that all squares are positive, and thus that the argument you gave stays true word for word if you start by assuming that $i\prec 0$. The point is really, that in $\mathcal C$ the number $-1$ is a square. In any ordered field, $1\ge 0$ and all squares are positive. It does not matter which of the two square roots of $-1$ you use, you'll get the same contradiction. 
