Why is it enough to have extended the numbers to include only one orthogonal imaginary axis? I am wondering in the context of roots of polynomials. I know that the orthogonality of imaginary axis w.r.t. real axis is not the only property of imaginary numbers but there is also the relation $i^2=-1$. But still I am not able to think why is the solution space of polynomials complete with expanding the uni-dimensional real axis to 2-dimensional Complex numbers?
Complex Numbers : Why stop at 2 dimensions
analytic-number-theorycomplex numbersnumber-systemspolynomialsreal numbers
Related Solutions
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 $x+\alpha y$.
We use the complex plane because it provides a very helpful visualization of complex numbers. The reason we can use it is because each complex number $z = a + bi$ is completely determined by the real numbers $a$ and $b$. In a way, thinking about a complex number as a point in $\mathbb{R}^2$ is just a different way of notating that number. There's no danger of confusion or ambiguity, because everyone can agree that a given point $(a,b)$ corresponds to the complex number $a + bi$, and vice versa. This is explained also in Atmos's answer.
I think it's worth mentioning that there is actually more than just this correspondence between the elements of each set, however.
We can also think about different operations that can be done with complex numbers and vectors in $\mathbb{R}^2$, and we can compare these. We know that complex numbers can be added together, as can vectors in $\mathbb{R}^2$. Complex numbers and vectors in $\mathbb{R}^2$ both add in the same way. Compare the following, for $a,b, c, d \in \mathbb{R}$
$$ (a + bi) + (c + di) = (a + c) + (b + d)i \\ (a,b) + (c,d) = (a + c, b + d) $$
You can see that to add two complex numbers you can just think of each complex number as a point in $\mathbb{R}^2$ and add these points as you normally would. The result is the point in $\mathbb{R}^2$ corresponding to the sum of the complex numbers.
We can also scale complex numbers (multiply them by a real constant), and also scale vectors in $\mathbb{R}^2$. Again, this scaling behaves in the same way in $\mathbb{C}$ as it does in $\mathbb{R}^2$. Let $a,b,k \in \mathbb{R}$ and compare again
$$ k(a + bi) = ka + (kb)i \\ k(a,b) = (ka, kb) $$
We see that to scale a complex number by a real constant, we can just think of the complex number as a point in $\mathbb{R}^2$, scale that point as we normally would, and the result is the point in $\mathbb{R}^2$ corresponding to the scaled complex number. If you'd like, the precise way of saying all of this is to say that $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic as $\mathbb{R}$-vector spaces. So you can be confident that the things you do with vectors (i.e. scalar multiplication, and vector addition) are the same in $\mathbb{C}$ and $\mathbb{R}^2$, so we are permitted to use this helpful visualization.
Edit: Here's an explanation of some extra behavior exhibited by $\mathbb{C}$ that is not compatible with the behavior of vectors in $\mathbb{R}^2$.
Above I did not explain the whole story of the multiplication operation. I explained how scaling by a real number works, but I did not mention what happens when you scale by an arbitrary complex number. Multiplication of complex numbers has a nice geometric interpretation when thinking of the elements of $\mathbb{C}$ as points on the plane. Each complex number $z = a + bi$ has associated to it a magnitude and an angle as measured counterclockwise from the positive $x$ axis. When multiplying complex numbers $z$ and $w$, the result is a new complex number with magnitude $|z|\cdot|w|$ and an angle that is the sum of the angles of $z$ and $w$. That is, multiplication of complex numbers corresponds to scaling and rotating on the complex plane. I explained a special case of this above when I talked about scaling by a real number. A real number has an angle of zero (it is on the $x$ axis), so multiplying by a real number only scales and does no rotation.
I mentioned that this operation is not compatible with a corresponding operation on $\mathbb{R}^2$. On one hand, this is the case because we are thinking of $\mathbb{R}^2$ as a vector space. And in a vector space, multiplication of two vectors need not be defined. So, this operation of multiplication is some sort of extra property possessed by the complex numbers that is not possessed by the vector space $\mathbb{R}^2$. This doesn't really change anything that I've already said, however. We can still think of complex numbers as points in $\mathbb{R}^2$. We just have to remember this extra rule for what happens to two points when we multiply them together.
And finally, some more detail if you still happen to be interested. We can define multiplication of elements in $\mathbb{R}^2$, and the natural way to do this is to say that for $a,b,c,d \in \mathbb{R}$
$$ (a,b) \cdot (c,d) = (ac, bc) $$
This makes $\mathbb{R}^2$ into a ring. This way of multiplying elements, however, is not the same as the way we multiply elements in $\mathbb{C}$ because in $\mathbb{C}$ we have the relation that $i^2 = -1$, which changes things slightly. The precise way to state this difference is to say that $\mathbb{R}^2$ and $\mathbb{C}$ are not isomorphic as rings. Again though, this often doesn't matter to us, because in the situations where we want to think of $\mathbb{C}$ as $\mathbb{R}^2$ we are thinking of their compatibility as vector spaces, not as rings.
Best Answer
There are several other useful extensions of $\mathbb{R}$. You can google quaternions (real dimension 4) and octonions (real dimension 8). Each time the dimension increases, we loose some property of $\mathbb{R}$. For example multiplication of quaternions is not commutative, and for octonions it is not associative. But there is a theorem of Frobenius which say that in some natural sense these are all possible extensions of $\mathbb{R}$.