# Order of pole and evaluating residue

$f(z)=\frac{ze^{iz}}{z^2+a^2}$

I need to determine the order of the poles and compute the residues.

To compute this, we were told to (in general), write $f(z)=\frac{g(z)}{(z-z_0)^m}$ , and choose $g$ so that $m$ is minimized (a natural number -- this is the order of the pole), my issue is two fold:

1)How do we know, in general, that we have picked a $g$ which minimizes the order.

2) How do I handle this particular function (and others with essential singuarities).

I know I may also write $f(z) =\frac{ze^{iz}}{(z+ia)(z-ia)}$, so I know the singularities are $\frac{+}{ }ia$ and an 'essential singularitiy' (at infinity). But not sure how to compute the residues....

EDIT: $\lim_{z\to ∞} \frac{ze^{iz}}{z^2+a^2}=\lim_{z\to ∞} \frac{ze^{iz}}{z^2}= \lim_{z\to ∞} \frac{e^{iz}}{z}=∞$

So I can't use the result for when the limit exists and is finite (zero or non-zero).

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See the section on calculating residues.en.wikipedia.org/wiki/Residue_(complex_analysis) – Christopher A. Wong Sep 12 '12 at 4:33
Yeah, I obviously know the simple formula involving factorials and derivatives, my issue is how do I know I've determined the correct order and also, I would like help evaluating this particular case. – Squirtle Sep 12 '12 at 4:36
No, you don't "obviously" know that formula - you didn't say that you knew it, so how would it be obvious that you knew it? Anyway, where do you get stuck trying to use that formula on this example? – Gerry Myerson Sep 12 '12 at 4:43
How do we know, in general, that we have picked a g which minimizes the order. Because I can't use the formula UNLESS I know the order, right? :( But I just don't know how to find the order.... sorry, for the silly assumption, btw. – Squirtle Sep 12 '12 at 4:44
@dustanalysis, See Niccolo's post about selecting $g$. You pick $g$ so that it doesn't vanish at the point you're considering. – Christopher A. Wong Sep 12 '12 at 5:04

## 2 Answers

The practical advice to determine the order of a pole $w$ of a function $f(z)$ is the following: write $f(z)$ in the form $$f(z)=\frac{g(z)}{(z-w)^m}$$ where $g(w) \neq 0$, then $m$ is the order of the pole $w$.

In your example, you already have the function $f(z)$ in the nice form, for example for $w=ia$ pick $g(z)=\frac{z e^{iz}}{z+ia}$ which is clearly non-zero when evaluated in $z=ia$, thus the order of the pole $z=ia$ is 1.

By the way, I assumed $a \neq 0$, try to work out on your own what happens when $a=0$.

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So if I understand this correctly.... I believe that when I want to do it for the other pole (-ia) then I will just let g(z)=ze^iz/(z-ia) and similarly see that it is of simple order. Great! So I merely need it to be analytic "enough" (i.e. not entire). Now,... I still don't know how to handle the essential singularity.... – Squirtle Sep 12 '12 at 5:10
Exactly, you need g(z) to be analytic in a neighborhood of the pole, not entire as in analytic on the whole complex plane. – Niccolò Sep 12 '12 at 5:18
Now, about the singularity at infinity, you see it has order 1 because once you factor $z$ from $f(z)$, the remaining function $g(z)=e^{iz}/(z-ia)(z+ia)$ tends to zero as $z$ goes to $\infty$. To calculate the residue, check the last formula here en.wikipedia.org/wiki/… – Niccolò Sep 12 '12 at 5:21
I'm confused... first off, the wiki page doesn't concern the convergence of g, but rather of f. Secondly, even if we consider your g(z) then I don't see how that tends to zero, as z tends to infinity. exp(z) > z^2. So the numerator tends to infinity "faster" and the limit does not exist, right? – Squirtle Sep 12 '12 at 6:03

I think that in general one has to be able to factor the function enough as to see the picture around some pole. In this particular case, as we have a rational function times an entire one, we can do:

$$z^2+a^2=(z-ai)(z+ai)$$

and from here we have two (or three, if you want to subdivide (2)) cases:

$$(1)\;\;\;a=0\Longrightarrow f(z)=\frac{ze^{iz}}{z^2}=\frac{e^{iz}}{z}\Longrightarrow \operatorname{Res}_{z=0}(f)=\lim_{z\to 0}z\,f(z)=\lim_{z\to 0}e^{iz}=e^0=1$$

$$(2)\;\;\;a\neq 0\Longrightarrow \operatorname{Res}_{z=\pm ai}(f)=\lim_{z\to\pm ai}(z\mp ai)\,f(z)=\lim_{z\to \pm ai}\frac{ze^{iz}}{z\mp ai}=\frac{\pm aie^{\mp a}}{\pm 2ai}=\frac{e^{\mp a}}{2}$$

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Thank you, but we still haven't covered the case when z=infinity for the second case (a not equal to 0). In other words, the essential singularity needs to be dealt with and the comments by @Niccolo haven't helped me here. – Squirtle Sep 12 '12 at 15:33
What $\,z=\infty\,$ are you talking about?? In (2) I deal with the general case $\,z=\pm ai\,\,,\,\,a\neq 0\,$...no infinity at all mentioned in my post. – DonAntonio Sep 12 '12 at 17:07
Yeah, I know you didn't mention infinity. That's precisely my issue. I have to consider when the function explodes as well, right? I.e. If z->infinity then f(z)->infinity. I feel like I'm not understanding something. – Squirtle Sep 12 '12 at 17:39