If complex numbers can be represented as vectors, why can't we define $2$-dimensional vector division just as complex division? If complex numbers can be represented as vectors, why can't we define $2$-dimensional vector division just as complex division?
Is there any inconvenient/incompatibility to this?
 A: The standard vector operations don't impose any preferred basis on the plane $\mathbb R^2$. Any unit vector is the same as any other unit vector. But if you identify $\mathbb R^2$ with the complex numbers $\mathbb C$ and introduce complex multiplication and division, then suddenly there is a special vector (identified with $1+0i$) that serves as the multiplicative identity. You'll have chosen a particular direction in space that this identity element points to. The whole space "knows" where the identity is, since for any nonzero vector $v$ we can take $v/v$. This is kind of a weird thing to do.
A: Complex numbers are used as a representation for 2 dimensional vectors. That is, they help simplify the mathematics and follow certain properties that apply to vectors. However, they are not the same thing. Hence, in the case of division, complex division has some meaning whereas vector division is meaningless. 
A: The usefulness of vector spaces is in their abstractness. Vector spaces capture features that are common to many more specific spaces such as function spaces, $\mathbb{R}^n$, sets of sequences, and sets of matrices. They help us discover things about these particular spaces by throwing out excess structure that might be obscuring our vision. 
An analagous example might be Euler's work on the seven bridges of Konigsburg (sp?). Legend has it that he had a map of Konigsberg, which contained relative positions of bridges and streets and buildings and waterways, and a scale, so that he could measure distances. But Euler found that all of this information just made the problem very complicated. He solved the problem by throwing away the map and instead considering the much more abstract concept of a graph (i.e. a network).
Likewise, when we start with any concrete example of a two dimensional vector space we might have a lot of structure in a addition to the vector space stucture. Consider the space of polynomials of degree 1. When we think of this space as a vector space we are purposefully disregarding certain aspects of the space so that we can more easily recognize the application of the many theorems of linear algebra for finite dimensional spaces. We could define a division so that, using the example in the comments, $(2+x)/(2x)=1/2 -x$, but this might not be meaningful for polynomials, or even worse, needlessly complicate things.
If instead we started with the complex plane, then this division would be meaningful, but we would probably only be calling it a vector space if we were trying to simplify things by momentarily ignoring the multiplicative group aspect of $\mathbb{C}$ in order to get to the root of what otherwise might be a complicated problem.
A: Yes, we could. But the objects that you can multiply and divide are called numbers, and the objects that you can multiply by scalars and can be arranged in linear combinations are vectors, although in some sense they are the same objects. That is, complex numbers and vectors of $\Bbb R^2$ are not the same things because their properties are different.
Just like a board over four legs: if you use it for writing is a desk, and if you use it for cutting wood is a workbench.
A: From my perspective, the answer is "Yes, we could define multiplication and division in $\Bbb R^2$ using $\Bbb C$, but it would be dishonest."
One major reason to like vector spaces because they're the objects upon which linear transformations are defined. These linear transformations, by definition, respect vector addition and scalar multiplication: Given vectors $x$ and $y$ in a vector space $V$, and a scalars $c_1, c_2$, then any linear transformation $T$ must obey
$$T(c_1x + c_2y) = T(c_1x) + T(c_2y) = c_1T(x) + c_2T(y).$$
If we did want to define multiplication and division, surely they too should respect the "linearity" of a linear transformation! That is, if we could multiply vectors $x$ and $y$, we would want
$$T(xy) = T(x)T(y), \quad \text{and} \quad T(x/y) = T(x)/T(y).$$
You can verify for yourself that many linear maps $T: \Bbb R^2 \to \Bbb R^2$ wouldn't respect this new multiplication. What should we make of them?
Further, the fact that $\Bbb C$ can be viewed as a $2$-dimensional vector space over $\Bbb R$ is really quite special; we shouldn't expect all $2$-dimensional vector spaces to have something like $\Bbb C$ to help them out.

If you're interested in such questions, you would probably be interested in (abstract) algebra. 
There you'll learn that vector spaces occupy a very specific niche: They're built up from an underlying abelian group (this is where vector addition takes place). It's enriched with scalar multiplication, thanks to a special kind of ring called a field, and this scalar multiplication 'plays nicely' with vector addition. That's all we want from vector spaces.
Vector spaces that do happen to have a sensible way to define vector multiplication are called algebras. So, the fact that $\Bbb C$ can be viewed as an $\Bbb R$-algebra really deserves the applause here, and isn't a good reason to define vector multiplication in $\Bbb R^2$. 
The moral of the story: Let vector spaces be themselves, and work with algebras if you want to multiply vectors.
A: This isn't a bad idea. Sometimes we do add structure to a space by extending it in order to "complete" the space with respect to some property. Working with the completion does simplify our lives, as long as we know we can interpret the final answer in terms of the original problem.
As an example, it is often hard to evaluate an integral over a domain in $\mathbb{R}$. We often want to evaluate integrals whose domain includes a singularity of the integrand. If we extend the integrand to the complex plane, which completes the reals in an algebraic sense, then we can integrate along a path that circumvents the singularity and end up with a real-valued anti-derivative. We also simplify our lives sometimes by writing things like $\mathrm{cos}(x)=\frac12 (e^{i x} + e^{-i x})$, or by expanding real valued functions into series of terms containing $e^{i\pi n x}$, $n\in\mathbb{Z}$. If we know that in the end we will have a real-valued answer then we benefit from temporarily adding this extra structure.
So the idea is a good, creative idea. It is possible that there is some application in which the lack of multiplication of vectors gets in the way. But if you temporarily add structure to solve a problem then you need a way to interpret the results in the original vector space. Also, in the spirit of the previous example, it might be more fruitful to define division by extending the original space to a larger space which contains a copy of $\mathbb{R}^2$ that completes $\mathbb{R}^2$ with respect to division.
