What does it tell me when I find that a polynomial has complex roots, except for the obvious fact that it crosses zero for these values?
To me it seems that the existance of complex roots must have some implication in the real domain as well, though I don't know what or why.
On
If a polynomial has complex roots, then in the real domain it will have less roots than would otherwise be "expected". When I say expected, we normally would think that a polynomial of degree $n$ would have $n$ roots, counting multiplicities, but this is only the case over an algebraically closed field (like $\mathbb C$, not $\mathbb R$). If you have $m$ complex roots then there will be $n-m$ real roots left (counting multiplicity).
For example if you have a quadratic polynomial in $x$ and $y$ with complex roots, then it is entirely in one half of the plane. Similarly a cubic in $x$ and $y$ with complex roots it will intersect $y=0$ once.
A complex root of a polynomial can have some significance itself when the roots of the polynomial have significance in general. One example that comes to mind where the roots of polynomials have a meaningful interpretation is in the field of dynamical systems.
Consider a matrix differential equation
$$ X' = AX, $$
where $A$ is a constant real $2\times2$ matrix and $X$ is a $2\times1$ vector of real functions. The system always has a zero solution $X = \mathbf 0$, and the eigenvalues of $A$, which are the roots of the polynomial equation
$$ \det(A - \lambda I) = 0, $$
determine how the flow of solutions to the differential equation behaves with respect to this zero solution. The expression $\det(A - \lambda I)$ is indeed a polynomial, called the characteristic polynomial of the matrix.
When both eigenvalues are real and positive the solutions flow away from the point $X = \mathbf 0$ as time increases. When the eigenvalues are both negative the solutions flow directly to $X = \mathbf 0$ instead. There are other cases too, but let's focus on these. Here are a couple plots to illustrate, the first with negative eigenvalues and the second with positive:
If there is one complex eigenvalue then there must actually be two complex eigenvalues since the coefficients of the characteristic polynomial are real, and the real parts of these eigenvalues are equal. The real part of these eigenvalues plays the same role as above: if it's positive the solutions tend toward the origin and if it's negative the solutions move away from it. What the imaginary part indicates is that there is some rotational quality to the flow of solutions. Instead of flowing directly to the origin the solution will rotate around it infinitely often as it approaches or moves away.
Here's a plot showing the flow of a system whose eigenvalues are $1\pm 2i$:
If the eigenvalues are purely imaginary (i.e. their real parts are zero) then the solutions no longer tend to or away from the origin. All that remains is their rotational component, and they travel in ellipses around the origin, never settling down, as in the following plot.
To summarize, the presence of complex roots of the characteristic polynomial tells us that the flow of solutions to the corresponding differential equation has a rotational quality which is not present when the roots are real. The roots being purely imaginary has the special significance in that the flow is purely rotational; the solutions no longer tend toward $\mathbf 0$ or $\mathbf \infty$.