[begin crackpot theory]
Mathematicians have committed a great blunder by thinking that a certain imaginary number is $i$ and another is $-i$, when it's really the other way around!!!!
[end crackpot theory]
If one were to take the position above, and try to rewrite all of mathematics consistently with the theory propounded above, the result would be that nothing at all would change. Which one is called $i$ and which is called $-i$ doesn't matter.
It does in some contexts make sense to pay attention to whether the real part of a complex number is positive. A Dirichlet series $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^s}$ converges if the real part of $s$ is greater than the real part of the abscissa of convergence of the series, and diverges if the real part of $s$ is less than the abscissa of convergence.
In principle we could do what you suggest -- take $\mathbb R^2$ and associate every point $(x,y)$ to the number $x+13k$. Though the trouble with that particular plan is that each number now represents many different points -- for example, $(13,0)$ and $(0,1)$ and $(26,-1)$ are now all associated to the number $13$. This means that we can't use the scheme for anything where we calculate a number and that number points to exactly one point in the plane.
We could, however, do something more general. Take some field that extends $\mathbb R$, pick some element $\alpha$ in that field, and then represent $(x,y)\in\mathbb R^2$ by $x+\alpha y$.
As it went for $13$, if we pick $\alpha\in\mathbb R$, then we get something where a number doesn't represent a unique point. Suppose, however, that we steer clear of that case, and furthermore that we end up in the lucky situation that every element of the field represents some $(x,y)$ in the plane.
Something wonderful happens then -- namely, we can then prove (though not in the space left for me in this margin) that the field we're using must be isomorphic to $\mathbb C$ -- in other words the field is essentially the complex numbers, just called something different. In particular, somewhere in the plane there is an $(x,y)$ whose corresponding number behaves exactly like $i$.
So we could actually have said: Pick some complex number $\alpha$, and let $(x,y)$ correspond to $x+\alpha y$. As long as $\alpha$ is not real, this will give us a perfectly good one-to-one correspondence between points and complex numbers.
Now, among all the possible choices of $\alpha$ it turns on that exactly when $\alpha=i$ or $\alpha=-i$ we get the additional nice property that multiplication by any fixed nonzero complex number will correspond to a transformation of the plane that takes geometric figures to similar geometric figures.
Having multiplication correspond to similarity transforms is a pretty nifty property, which is a reason to prefer the representation $x+iy$ over other possible $x+\alpha y$.
Best Answer
There is no way to order complex numbers, in a way that preserves operations in a sensible way. The precise term is ordered field; among their properties, $x^2 \ge 0$ for every $x$. Since $i^2=-1$, we would need $-1\ge 0$, which is impossible.
However, if you want to measure distance, you can do that with a norm. In the complex numbers, this is calculated as $|a+bi|=\sqrt{a^2+b^2}$. Hence, it is correct to say that $2i$ is twice as far from the origin as $i$ is.