# [Complex Analysis] Determining order of a pole.

1. May 28, 2012

### Je m'appelle

I've been studying the residue theorem and I've been having some difficulty with classifying singularities.

For example, let's use the function

$$f(z) = \frac{1}{z sinz}$$

I know it has two singularities, one at z=0 and the other at z=2kπ for k ={0,1,2,..}, I don't know what kind of singularities they are so I'll rewrite it in terms of Laurent series

$$f(z) = \displaystyle{ \frac{1}{z} \frac{1}{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2n+1}} }$$

I also know I could rewrite it in terms of $$z^{2n+2}$$ and realize that at n=0 there's a singularity (due to z=0?) but how do I go on about identifying them (essential, removable or pole)? And how could I analyze the other singularity (due to z=2kπ), should I rewrite the series in terms of (z-2kπ) or something?

PS: In the title I stated "determining order of a pole" because I know in this particular case the singularities are poles as I took this function as an example out of a "determine the order of the poles" problem. Anyway, disregard the title and sorry for the confusion.

Last edited: May 28, 2012
2. May 28, 2012

### tt2348

The residue theorem works based on your domain. IE if your singularity is contained in the closed curve. Other things to consider are winding numbers ( think branch points) in regards to the singularities.

3. May 28, 2012

### micromass

Ok, let's do it first for z=0.

1) What is $\lim_{z\rightarrow 0} f(z)$?? If this limit exists, then the singularity is removable. If not:

2) What is $\lim_{z\rightarrow 0} zf(z)$?? If this limit exists, then there is a pole of order 1. If not:

3) What is $\lim_{z\rightarrow 0}z^2f(z)$?? If this limit exists, then there is a pole of order 2. If not:

4) etc.

So, which one will it be?

4. May 29, 2012

### Je m'appelle

I can see that for z=0 it's a pole of order 2 due to it's taylor series

$$f(z) = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2n+2} \right)^{-1}$$

as for n=0 we have our first singularity $z^{-2}$ which I can verify it as a pole by it's definition "The coefficients $a_n$ are zero for n<-k and $a_{-k} \neq 0$".

Also by your method, it's clear that

$\lim_{z\rightarrow 0} f(z) = \lim_{z\rightarrow 0} zf(z) = +\infty , \ \lim_{z\rightarrow 0}z^2f(z) = 1$

But what about z=kπ? Can I follow the same logic of Taylor series? But how should I proceed?

====================================================================================
EDIT: I was thinking something along the lines of

$$f(z) = \frac{1}{sinz} \frac{1}{k\pi + (z - k\pi)}$$

$$f(z) = \frac{1}{sinz} \frac{1}{k\pi \left( 1 + \frac{(z-k \pi)}{k \pi}\right)}$$

and since

$$\frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-1)^n x^n$$

then

$$f(z) = \frac{1}{sinz} \left( \sum_{n=0}^{\infty} (-1)^n(k \pi)^{1-n} (z - k \pi)^n \right)^{-1}$$

and the other singularity would be a pole of order 1 for z=kπ, k={1,2,3,...}, at n=1?

But is this correct? Should I have also expanded the sine or not?

=====================================================================================

I understand that but what if it's not stated? Or does it always have to be stated?

What do you mean?

Last edited: May 29, 2012
5. May 29, 2012

### micromass

As for the poles at $k\pi$. You have to find the least n such that

$$\lim_{z\rightarrow k\pi} (z-k\pi)^nf(z)$$

exists.

One method is by expanding sin(z) in it's Taylor series around $k\pi$.

6. May 29, 2012

### Dickfore

Find the limit:
$$\lim_{z \rightarrow 0} \left( \frac{z^{n}}{z \, \sin z} \right)$$
for various positive integers n.

7. May 29, 2012

### algebrat

division is evil, make it go away.

That is, for the ratio of two laurent series, if you want to find exactly which Laurent series it results in, assume one, that is, write

$$\frac{\sum a_j z^j}{\sum b_k z^k}=\sum c_\ell z^\ell.$$

Then

$$\sum a_j z^j=\sum b_kz^k\sum c_\ell z^\ell.$$

So now we have multiplication, which is a little hard, but less hard than division. Division always sucks, for instance, what's easier, product rule or quotient rule?

This multiplication trick is great, you can prove the result micromass gave you, and many more great results about rational functions in complex analysis. Like the residue trick for f/g, where f and g satisfies some conditions. This trick will show you why those conditions work.

For $\frac{1}{z\sin(z)}$, we try

$$1=z\sin(z)(\sum_{\ell=m}^\infty c_\ell z^\ell)=(z^2+\cdots)(c_m z^m+\cdots)$$

This tells us that $m=-2$, quickly and convincingly.

Of course this is the trick at the origin, so we can shift it to make the argument elsewhere. But it's good to able to just look at it and see the roots are poles of order 2 at the origin, and 1 elsewhere. Still should check though.

8. May 29, 2012

### tt2348

Your domain generally dictates the form your laurent series would take . I'm assuming the closed curve you'd be evaluating over would only go around once, so don't worry about winding numbers (the amount of times the curve goes around a single point). But in regards to the residues. keep in mind that (-1)-k sin(z)=sin(z+k*π). for instance, if we wanted to consider a domain around... |z|=3, wed only be concerned with z=0, but the more singularities encompassed by the curve, the more residues we add.
Using the periodic nature of sin(z), we'd have sin(z)= Ʃanzn=(-1)kƩan(z+kπ)n
that is , your coeffiecients for the laurent series for each singularity only changes by switching between + or -

Last edited: May 29, 2012