Are imaginary numbers really incomparable?
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.
There is a good reason to avoid defining "orderings" like $i > -i$, or any similar comparisons, even if you don't care about ordered fields. That is: there is no way to distinguish between $i$ and $-i$, except for the symbols used to denote them. $i$ is defined as a solution of $x^2 +1 = 0$. But there are two solutions, so how do you decide which one gets the privilege to be bigger than the other?
If you take any sentence in the language of complex numbers in which $i$ occurs, and replace it everywhere with $-i$, then the truth of the sentence doesn't change. That isn't true anymore if you introduce definitions like $i > -i$.
We plot $i$ before $2i$ and so on purely by convention (and maybe that our brains are acquainted to working with numbers from smaller magnitudes to larger magnitudes). On a computer screen, for example, $2i$ would have lower $y$ coordinate than $i$ and thus is plotted first.
There is no way of imposing a total ordering on $\mathbb C$ that would give you the same nice properties as one on $\mathbb R$, for example, having $x^2 \geq 0$ for all $x$, is not possible.
You can compare the magnitudes of complex numbers. That is, their absolute values. Distance in this case does matter because distance is a real number and is thus comparable. It is just that the complex points themselves are not comparable.