Effective equality (I believe this belongs to computational algebraic number theory, but if additional/different tags are better, let me know.)
Consider the cubic equation $$x^3-3x^2+x-3=0.$$ We have that $x=3$ is a root. On the other hand, one can use one of the algorithms for solving cubics (the methods followed by Tartaglia or others), and find a different looking expression, namely
 $$ x= 1 +\root 3\of{2+\frac{10}{3\sqrt3}} + \root 3\of{2-\frac{10}{3\sqrt3}}. $$
In this example, one can check that the numbers are the same, by noting that $$ \root3\of{2+\frac{10}{3\sqrt3}}=1+\frac1{\sqrt3}, $$ but this leads to my question:

Suppose that $\alpha$ and $\beta$ are complex numbers with explicitly given expressions, and that $\alpha$ and $\beta$ belong to some finite extension of ${\mathbb Q}$. 

By "explicitly given", one can mean that finitely many irreducible polynomials $p_i$ with integer coefficients are given, and that $\alpha$ and $\beta$ can be obtained from roots of these $p_i$ by radicals (as the expressions above), perhaps with additional information on which of the roots of the $p_i$ one is considering ("the smallest positive root," "the one whose imaginary part is positive and second largest," ...). If you see a more precise way of making sense of this, that is fine as well. 
In the example I was discussing, if I had not noticed that the cubic root simplified as it did, a reasonable way of convincing myself the two numbers were actually the same would have been to compute them. Using lots and lots of digits. 

Compute numerical approximations to $\alpha$ and $\beta$ using some reliable CAS. Can we explicitly find an a priori (and feasible?) bound for the number of digits one needs to compute to ensure that if both approximations coincide, then in fact $\alpha=\beta$?

The bound most likely would depend not just on the degree of the extension of ${\mathbb Q}$ where we can find $\alpha$, $\beta$, and all the radicals that make them up, but I imagine the degree is part of it.
(Somebody told me the LLL algorithm is perhaps the way of doing this, but this is all new to me.)
 A: After a discussion in the comments with Qiaochu Yuan, I think we both agree that this would be a suggested algorithm in practice : 
You are given two algebraic numbers $\alpha$ and $\beta$. Compute their respective minimal polynomials. If you get two distinct polynomials, you're done ; distinct minimal polynomials have distinct roots. If they are the same, compute numerically all the roots, and also compute
$$
\delta \overset{\text{def}}{=} \underset{\text{roots}}{\min} \{ \|r_i - r_j\| \}
$$
where $i$ and $j$ just mean "go over all distinct pairs of roots". This will be a (numerical) upper bound that will tell you if $\alpha$ and $\beta$ are distinct : now if you evaluate $\alpha$ and $\beta$ up to a precision of $\delta$ you should be able to tell if they are distinct or not.
Hope that helps,
A: Suppose that $a$ and $b$ are given by your expression, or something like it. Then we can construct formulas $F_a(x)$, $F_b(y)$ in the first-order language $\mathcal{L}$ with constant symbols $0$ and $1$, and binary function symbols $+$ and $\times$ such that $F_a(x)$ "says" that $x=a$, and $F_b(y)$ says that $y=b$. 
One can do this with expressions that are substantially more general than the ones you envisage, including expressions that are defined piecewise.
Let $\varphi$ be the sentence 
$$\forall x\forall y((F_a(x)\land F_b(y))\longrightarrow (x=y)).$$
Then $\varphi$ "says" that $a=b$.
Now use Tarski's decision procedure for the first-order theory of real-closed fields (or for algebraically closed fields of characteristic $0$, if appropriate) to decide the truth of $\varphi$.  (Since the original Tarski proof, there have been substantial improvements in the decision procedure, and even some implementations.)
Remark: Using a "general" decision procedure to solve a limited class of problems almost guarantees inefficiency. So there remains the very real problem of implementation for problems of the specific type you are interested in. 
A: This may help. Suppose $r$ is a root of polynomial $p(z)$, and you can get bounds $|p'(r)| \ge \alpha > 0$ and $|p''(z)| \le \beta$ for $|z−r| \le \delta$. Then if $2 \alpha > \beta \delta$, there are no other roots of $p(z)$ with $|z−r| \le \delta$.
