To solve polynomials, the answer is no. The Fundamental Theorem of Algebra says that every (non-constant) polynomial, of degree $n$, with complex coefficients has $n$ roots (including repeated roots) in $\mathbb{C}$.
As a result of this, the set of complex numbers, unlike the set of real numbers, is algebraically closed, which means that we cannot 'escape' $\mathbb{C}$ using any elementary operations, like $+, - , \times, \div, \sqrt{}, e^{...}$ etc. So, in this sense, we don't really 'need' to extend the complex numbers.
However, there do exist hypercomplex numbers, like the quaternions, $\mathbb{H}$, which, instead of using just one imaginary unit $i$, use three: $i, j, k$, each satisfying:$$i^2=j^2=k^2=ijk=-1$$
Quaternions are used in modelling 3D vectors, and have a lot of use in 3D mechanics.
A useful property of a quaternion is that it does not satisfy commutativity: e.g. $i\times j \neq j \times i$.
N0w, if you still want to go further, you can explore the octonions, $\mathbb{O},$ which are an extension of the quaternions.
Octonions have 7 imaginary units: $e_1, e_2, ...e_7$.
Octonions have far fewer (that I know of, at least) practical uses that quaternions.
Octonions also have the interesting property that they lack associativity: e.g. $x\times(y\times z) \neq (x \times y) \times z$ (for $x,y,z \in \mathbb{O}$).
And, finally, at least as far as mathematical interest has gone, we have the sedenions, $\mathbb{S}$.
These are an extension of the octonions and have 15 imaginary units: $e_1,e_2,e_3,...,e_{15}$. The special property about the sedenions is that they have zero-divisors (meaning that there exist non-zero sedenions $x$ and $y$ such that $xy=0$). Interesting, this property, isn't it? Not particularly, intuitive, to say the least.
Now, I'll leave as an exercise for you to find out about the applications of hypercomplex numbers, and how to multiply them (hint: Google 'Fano plane mnemonic', which explains how to multiply octonions- this same idea can be extended to the sedenions).
See http://en.wikipedia.org/wiki/Hypercomplex_number
There can be complex non-real numbers in the denominator. It just happens that if you have an expression of the type$$\frac{a+bi}{c+di},$$with $d\neq0$, if you convert it to$$\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac+bd+i(bc-ad)}{c^2+d^2},$$you get, in general, a more readable expression.
But you are not required to do this. Complex Analysis has lots of formulas that begin with $\frac1{2\pi i}$ and, as far as I know, nobody ever suggested that one should use $-\frac i{2\pi}$ instead.
Best Answer
There are several ways to introduce the complex numbers rigorously, but simply postulating the properties of $i$ isn't one of them. (At least not unless accompanied by some general theory of when such postulations are harmless).
The most elementary way to do it is to look at the set $\mathbb R^2$ of pairs of real numbers and then study the two functions $f,g:\mathbb R^2\times \mathbb R^2\to\mathbb R^2$:
$$ f((a,b),(c,d)) = (a+c, b+d) \qquad g((a,b),(c,d))=(ac-bd,ad+bc) $$
It is then straightforward to check that
With this construction in mind, if we ever find a contradiction involving complex number arithmetic, this contradiction can be translated to a contradiction involving plain old (pairs of) real numbers. Since we believe that the real numbers are contradiction-free, so are the complex numbers.