Find a polynomial that has two algebraic numbers as a root If you have two algebraic numbers $\alpha$ and $\beta$ with two polynomials $u(x)$ and $v(x)$ such that $u(\alpha)=0$ and $v(\beta)=0$ you can, for example, find out a polynomial $q(x)$ that has $\alpha+\beta$ as a root using the multivariate resultant:
$q(x) = res_y(u(x-y),v(y))$
then:
$q(\alpha+\beta) = 0$
Now my question is:
If $\alpha$ is the kth root of $u(x)$ and $\beta$ is the nth root of $v(x)$, is there a way to find out which root of $q(x)$ is $\alpha+\beta$ ?
 A: I am not aware of any direct way to calculate the "index" of $\alpha+\beta$ as a root of $q$. However, you can isolate the roots of $u$, $v$ and $q$, to get disjoint intervals $(a^-_i,a^+_i)$ with rational endpoints s.t. $\alpha \in (a^-_k,a^+_k) =: I_\alpha$ and similarly for $\beta$ and $\alpha+\beta$.
Then refine the intervals $I_\alpha$ and $I_\beta$ around $\alpha$ and $\beta$ and the "candidate intervals" containing the roots of $q$ until $I_\alpha + I_\beta$ overlaps with exactly one of the "candidate intervals" of $q$. This is an isolating interval of $\alpha+\beta$, and it allows you to compute its "index" among all roots of $q$.
Depending on the method you use (e.g., for most algorithms based on Descartes' rule of signs), you have to make the polynomials $u$, $v$ and $q$ square-free first. Also, if you do a square-free factorization of $q$, the isolating interval for $\alpha+\beta$ as computed above will contain a root of exactly one of the factors, which can easily be found by computing the sign of the factors at the endpoints of the interval. This factor will be the minimal polynomial of $\alpha+\beta$.
By the way, I was assuming that you deal with real roots (since you have that order on the roots). But the same procedure will also work for complex roots; just replace the concept of an isolating interval on the real line by isolating discs in the complex plane.
[edit]
Sorry, I did not recognize that the case of complex roots was specifically asked for.
With respect to your specific ordering, the problem is that it is not trivial to compare the argument of algebraic numbers (in particular because the argument is not algebraic itself).
I think the problem would be somewhat easier if the magnitude were the dominant key in the comparison.
Nevertheless, both imaginary and real part of an algebraic number are again algebraic, so you can decide whether the argument is zero, and you can represent the magnitude.
To compare two algebraic numbers $0 \ne \alpha \ne \beta \ne 0$ (those conditions can be checked "easily") according to your definition, first check whether their imaginary part has different sign.
If so, they lie in different halfplanes, and sufficient refinement will show which one has the smaller argument.
Similarly, if both imaginary parts are zero, both $\alpha$ and $\beta$ are real, and it the comparison can be done by refining against zero.
Otherwise, compute $\gamma = \alpha / \beta$, which again yields an algebraic value, and compute its imaginary part.
If it is zero, then $\arg\gamma = \arg\alpha - \arg\beta = 0$, hence their argument is the same.
(Note that the case $\arg\gamma = \arg\alpha' - \arg\beta' = \pi/2$ cannot occur, since it implies that exactly one of $\alpha$ or $\beta$ is a positive real and the other negative real, and this case has been dealt with already.)
However, if the imaginary part is not zero, you know the argument of $\alpha$ and $\beta$ differs, and again sufficient refinement of both numbers will allow a comparison.
Note that it is not sufficient to blindly start refining in the hope that you will be able to distinguish the argument - you need to know in advance that the procedure will terminate, and that's why the "boilerplate" checks need to be there. Also note that the computations involved in this approach will be very inefficient; essentially, you will need to compute the minimal polynomial (or at least a representing polynomial) for all algebraic values occuring throughout the computation.
