Can there be a third type of unit other than 1 and i? I know the question sounds vague, so let me try to explain what is on my mind. When we imagine the complex plane, one axis represents the real part and the other axis represents the imaginary part. You cannot represent the real part with the imaginary part and vice versa. (As far as I know, this represents basis vectors.) What can be an example to the third axis to these to axes, such that a point can only be represented as a linear combination of the basis vectors in such way:
$$
P(a,b,c)=a1+bi+cj \ \ \ (a,b,c \in \mathbb{R})
$$
Can it be a new set of numbers?
 A: Yes. In fact these generalized 'fields' of complex numbers have a pretty active fields too.
The first one is the Quarternions which have 1 'real' vector parts and 3 others. But these are not commutative in multiplication.
There are further levels too, like Octonions and Sedenions. But these aren't even fields (you can't divide on them properly). Thus the more you try to generalize these concepts into an 'algebraically closed' form the less algebraic they will become. At which point you might as well just use vectors and classical linear algebra.
Their applications are amongst Quantum Field Theory, generalized description of Electromagnetic fields, the specific mathematical field that tries to discover the noncommutativity of Quarternions(thus making chirality a measurable difference) is called Geometric Algebra or its older name, Clifford Algebra.
Notice how to be able to perform a vectorial product you need at least three dimensions.
$$\vec{u}\times\vec{v}=\vec{x}$$
The only other number of dimensions where the cross product is well-defined in simple vectorial notation is 7 dimensions.
To be able to calculate the cross product in any n-dimensions Grassmann introduced the wedge product, which differentiates between a product of vectors and a product of a vector and a scalar.
In classical linear algebra the product of two vectors leaves you with a vector in the $3$ dimension which is perpendicular to the plane which was created by the two vectors.
Meanwhile the product of a vector and a scalar is just our original vector which's length was changed by the scalar.
If we have a vector how do we know if it was created from two vectors or just one?
The answer: let's define a product
$$\vec{u}\wedge\vec{v}=\frac{1}{2}(\vec{u}\vec{v}-\vec{v}\vec{u})$$
where $\vec{u}\vec{v}$ is the scalar product of the two vectors.
If we switch them and take their product:
$$\vec{v}\wedge\vec{u}=\frac{1}{2}(\vec{v}\vec{u}-\vec{u}\vec{v})=-\frac{1}{2}(\vec{u}\vec{v}-\vec{v}\vec{u})=-\vec{v}\wedge\vec{u}$$
it's clear that they satisfy the anticommutativity of the original cross product.
The quantity that we get when we take the wedge product is called a bivector which is the set of two vectors and a rotational component between them.
Why is this so good? Well, it turns out this concept generalizes really well to $n$-dimensions and we can do algebra on them while not having to deal with visual downsides of the cross product.
A: In order for one new unit to exist, we have to answer this question:

what's $i\times j$?

the answer to this question has to have a particular property: the result can't be the same as any other question with either $i$ on the left or $j$ on the right and any of $1, -1, i, -i, j, -j$ filling in the other side.  so it can't be: $i\times 1 = i$, or $i\times -1 = -i$, or $i \times i = -1$ or $i \times -i = 1$, and it can't be $1 \times j = j$ or $-1 \times j = -j$, so ... we're kinda stuck.
Fortunately Hamilton went "wait what if we added yet another unit, $k$?" and then we ended up with the quaternions, which get a lot of use because unit quaternions are a double cover of 3d rotation.  There are others too, but quaternions are the ones that get the most press.
