2

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!

Daniel Cherno
  • 23

2 Answers2

1

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.

Michael Seifert
  • 9,477
0

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.

Daniel Cherno
  • 23