Ok, what exactly is it that you want? A procedure that gives the Laurent series (about 0) for the square root of any even degree polynomial? Of course even degree polynomials have Laurent series for sufficiently large annuli, because the x^2n term dominates and has a single-valued square root (2, in fact). On the other hand, if your polynomial has odd degree you won't have a Laurent series for large annuli - your only hope is for small annuli if the other terms let you behave nicely enough somewhere that you have a single-valued square root, which isn't ensured in general.
Anyways, wouldn't any reasonable square root algorithm be analogous to the division algorithm, and therefore calculate higher order terms first and then calculate lower order terms, which would pretty much destroy all hope when you have infinitely many terms with positive degree?
Of course, if you just look in a neighborhood around some point z_0 your polynomial p is nonzero, you can get a perfectly nice Laurent expansion for sqrt(p(z)) - in fact a Taylor series expansion. And since Taylor series are unique, you should just get the Taylor series for sqrt (about p(z_0)) applied to p(z) (i.e. plug in p(z)^n for z^n, expand it out, and regroup terms so you again get a power series). Which means that if p(z) is already a square in C[z], you should get a lot of cancellations of terms in the Taylor series and be left with it's square root (since Taylor series are unique). That's what I meant by substituting it into the series for sqrt(x).
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson