Is it possible, for an arbitrary polynomial in one variable with integer coefficients, to determine the roots of the polynomial in the Complex Field to arbitrary accuracy? When I was looking into this, I found some papers on homotopy continuation that seem to solve this problem (for the Real solutions at least), is that correct? Or are there restrictions on whether homotopy continuation will work? Does the solution region need to be bounded?

This argument is problematic; see Andrej Bauer's comment below. Sure. I have no idea what an efficient algorithm looks like, but since you only asked whether it's possible I'll offer a terrible one. Lemma: Let $f(z) = z^n + a_{n1} z^{n1} + ... + a_0$ be a complex polynomial and let $R = \text{max}(1, a_{n1} + ... + a_0)$. Then all the roots of $f$ lie in the circle of radius $R$ centered at the origin. Proof. If $z > R$, then $z^n > R z^{n1} \ge a_{n1} z^{n1} + ... + a_0$, so by the triangle inequality no such $z$ is a root. Now subdivide the disk of radius $R$ into, say, a mesh of squares of side length $\epsilon > 0$ and evaluate the polynomial at all the lattice points of the mesh. As the mesh size tends to zero you'll find points that approximate the zeroes to arbitrary accuracy. There are also lots of specialized algorithms for finding roots of polynomials at the Wikipedia article. 


Homotopy continuation method is good for finding all COMPLEX solutions to arbitrary accuracy, and it is implemented in the Numerical Algebraic Geometry package in Macaulay 2, for example. The method is more general. It can solve a system of polynomial equations in many variables. In fact, it is a more difficult problem to find all REAL solutions WITHOUT finding all complex solutions. From what I understand, the solution region does not need to be bounded for homotopy continuation to work. You can also "projectify" your problem if necessary, so that you don't have to worry about homotopy paths going off to infinity. Some methods assume that the solutions are all simple, but there're ways to work around it. One is the method of "deflation". 


For univariate polynomials you should look at "An Efficient Algorithm for the Complex Roots Problem" by Andy Neff and John Reif http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=5E9156BAF80D8D6AEDCA2F42C11AB4B2?doi=10.1.1.33.3353&rep=rep1&type=pdf 


The wikipedia article http://en.wikipedia.org/wiki/Rootfinding_algorithm gives links to many different methods for finding roots of polynomials. (Start at the section entitled "Finding roots of polynomials".) Many of the methods are incomparable, in the sense that they work faster or slower than others depending on the specific polynomial. 


One of the semirecommended ones for finding roots in the complex plane is Laguerre's method, which for some reason is not included in the Wikipedia article on rootfinding. http://en.wikipedia.org/wiki/Laguerre's_method The reason I know of this is a colloquium lecture long ago by Steven Smale on the complexity of Newton's method, during which William Kahan stood up and held forth on why Newton's method was worthless and Laguerre's was much better. I cannot tell whether you insist on finding all roots to high accuracy. One could perhaps divide out by $(x  r_k)^{n_k}$ each time a root $r_k$ with multiplicity $n_k$ is found, and search for roots for the new polynomial, using those results as seed values for finding accurate roots using the original polynomial. 


This can be done. Check this article by Hubbard, Schleicher, and Sutherland, entitled "How to find all roots of complex polynomials by Newton's method". 


Although it's not specific to polynomials with integer coefficients, have a lot at "Computing the Zeros of Analytic Functions". 


At least for real roots it can be completely solved by bracketing zeroes with Sturm sequences. 


You have already seen McNamee's excellent bibliography on polynomial rootfinding methods? Personally I have a preference for the "simultaneous iteration" methods (of which DurandKerner and EhrichAberth are two of the simplest and most wellknown); all you need to start from is a set of points equispaced around a circle in the complex plane (as to the radius of this circle, there are a number of suggestions in the literature; alternatively, formulas in Marden's "Geometry of Polynomials" might be of use here). 


A completely ineffective theoretical method goes as follows: Write $f(z)$ as $f_r(x,y)+if_i(x,y)$ where $f_r,f_i\in \mathbb R[x,y]$ are real polynomials in the real and complex part $x,y$ of $z=x+i y$. Compute a Groebner basis of the ideal $(f_r,f_i)$ with respect to an order which eliminates one of the variables in the first element of the basis and use real techniques (based on Sturm sequences) to compute, say, the real parts of all solutions. Use another element of the Groebner basis (or again real techniques) to compute the corresponding imaginary part and test for multiplicities (which can be avoided by computing first gcd$(f,f')$). Completely useless (and equivalent) variation: Study the intersection giving the zeroes of $f$ of the two real curves determined by $f_r$ and by $f_i$. 

