How to evaluate poles of fractional order (that do not lie on branch cuts!) I would appreciate some guidance. I am trying to evaluate the residue of a function that looks like this:
$f(z)=\frac{g(z)}{p-\sqrt z}$
at $z=p^2$ in the complex plane. $p>0$ is real, so I do not believe this pole lies on the branch cut, which I choose to start at the origin and move along the negative real line. Furthermore, the function $g(z)$ is analytic around $z=p^2$, but I assume it, and its derivatives, will feature in the final answer.
How do I go about determining the order of this pole, for instance? Mathematica claims it's order 1. Do I Taylor expand the root around the pole and just use the linear term?
Finally, the ultimate question, which I hope to answer once I understand the intermediate, is what happens in this case:
$f_k(z)=\frac{g(z)}{(p-\sqrt z)^k}$
for $k\in N$ some positive integer.
Thank you!
 A: The easiest way to see what's going on (for both of your function) is to rewrite $f(z)$ as
$$
f(z) = \frac{g(z)}{p - \sqrt{z}} \frac{p + \sqrt{z}}{p + \sqrt{z}} = -\frac{g(z) (p + \sqrt{z})}{z - p^2}.
$$
Since $p + \sqrt{z} \neq 0$ in a neighborhood of $z = p^2$, this is a legitimate move.  In this form, it's fairly obvious that the residue of the function is $- 2 p g(z)$.  Similarly, the higher-order functions you're ultimately interested in will have poles of order $k$.
Just to be sure that everything's on the level, let's imagine integrating $f(z)$ around a small circular contour of radius $\epsilon$ centered at $p^2$.  The integral around this contour should be the residue of $f(z)$.
Along this contour, we have $z = p^2 + \epsilon e^{i \theta}$, where $\theta$ goes from $0$ to $2 \pi$.  Thus,
$$
\oint f(z) dz = \int_0^{2 \pi} \frac{g(p^2 + \epsilon e^{i \theta})}{p - \sqrt{ p^2 + \epsilon e^{i \theta}}} (i \epsilon e^{i \theta} \, d \theta) 
$$
In the limit $\epsilon \to 0$, the integrand approaches
$$
\frac{g(p^2 + \epsilon e^{i \theta})}{p - \sqrt{ p^2 + \epsilon e^{i \theta}}} i \epsilon e^{i \theta} \approx \frac{g(p^2) + \epsilon e^{i \theta} g'(p^2)}{p - (p + \frac{\epsilon}{2 p} e^{i \theta})} i \epsilon e^{i \theta} \approx - 2ip g(p^2)
$$
and since the function is "well-behaved" near $z = p^2$, the limit of the integral is the integral of the limit and we have
$$
\oint f(z) dz = (2\pi) (- 2ip g(p^2)) = - 4 \pi i p g(p^2).
$$
Thus,
$$
\mathrm{Res}(f(z), z = p^2) = - 2 p g(p^2),
$$
as we found above.
A: As Michael points out, I should redefine as
$f(z) = \frac{g(z)\cdot(p+\sqrt z)}{p^2-z}$
and in the more elaborate example:
$f_k(z) = \frac{g(z)\cdot(p+\sqrt z)^k}{(p^2-z)^k}$
And the orders appear clear as day. Thanks for a quick resolution.
