As others have mentioned, Hamilton came up with the quaternions in a successful attempt to describe 3-D rotations. Complex multiplication can model 2-D rotation. Quaternion multiplication can be used to model 3-D rotation.
I didn't see any answers to the original question though:
Are there some equations in complex that do not have solutions?
Well, I'm not sure what you'd call an equation exactly. Certainly the equation: 1/x = 0 has no solutions in the integers, rationals, reals, complexes, or quaternions. But...
There are polynomials with integer coefficients that have non-integer solutions. There are polynomials with rational coefficients that have non-rational solutions. There are polynomials with real coefficients that have non-real solutions.
However, every polynomial with complex coefficients has a complex solution.
Polynomials aren't quite as useful for quaternions since quaternions aren't commutative. You may have that (ax)<sup>2</sup> isn't equal to a<sup>2</sup>x<sup>2</sup>. This makes polynomials messy. Rather than being able to write something like: a + bx + ex<sup>2</sup>, you may have to write:
a + bx + xc + ex<sup>2</sup> + x<sup>2</sup>f + gxhx + xsxt + uxvxw.
That can be grouped a little better than that. But, you get the point.
People often focus on subsets then of left or right polynomials.... so a + bx + ex<sup>2</sup> is a left-polynomial (or is it right?) and a + xb + x<sup>2</sup>e is a right-polynomial (or is it left?). Every left polynomial with quaternion coefficients has a quaternion solution. Every right polynomial with quaternion coefficients has a quaternion solution.
There are some polynomials with quaternion coefficients that are neither right nor left (they are mixed...) which have no solutions in the quaternions.
The rationals are a ring-extension of the integers. The reals are a field-extension of the rationals. The complex numbers are a field-extension of the reals. The quaternions are a field-extension of the complexes.
There is no field-extension of the quaternions that provides solutions to the mixed polynomials. At least that's what this claims.
http://www.math.niu.edu/~rusin/known-math/95/quaternion.eq.
But, it's not clear to me what trouble comes from adding in those solutions. I'm thinking that what happens is that the whole thing collapses and you can show that everything equals zero if you append those solutions.