# Evaluating Residues

1. Sep 12, 2008

### Lyuokdea

I'm looking at an integral which in part involves finding the residue of $$\frac{1}{1+z^{n}}$$ at $$z=e^{\frac{i\pi}{n}}$$

I thought the general method for residues was to find the 1/z term in the Laurent series, (which i'm not particularly sure how to do in this case), however, the answer provided does:

$$\frac{1}{\frac{d}{dz}(1+z^n)} = -\frac{1}{nz^{n-1}}$$

evaluating this at $$z=e^{\frac{i\pi}{n}}$$, they obtain:

$$Res=-\frac{e^{\frac{i\pi}{n}}}{n}$$

Why is this a legitimate method, for solving residues, can you always employ this method (it seems much easier), are there any other important methods for determining the residues of various things?

Thanks,

~Lyuokdea

2. Sep 13, 2008

### Lyuokdea

sorry to do the bump, but anybody have insight on this? I have a big test monday, and this is continuing to trip me up

~Lyuokdea

3. Sep 13, 2008

### Hurkyl

Staff Emeritus
What methods and formulas do you know for calculating residues?

Last edited: Sep 13, 2008
4. Sep 13, 2008

### Lyuokdea

the main method i'm aware of was to find the Laurent series, and then find the coefficient of the 1/z terms...but, i'm not exactly sure how to do that in this case...working from a taylor series around $$z=e^\frac{i\pi}{n}$$, you would get infinities because the denominator is always 0...

if you had a set number for n, say n=2, it's easy because you factor that as (i+z)(-i+z), and you do partial fractions to get A and B coefficients, you could possibly work with a similar process for n, (i+z^n/2)(-i+z^n/2), but that doesn't give you the first order denominator you want.

~Lyuokdea

5. Sep 13, 2008

### jostpuur

There exists Taylor series

$$1 \;+\; z^n \;=\; a_1(z - e^{i\pi/n}) \;+\; a_2 (z-e^{i\pi /n})^2 \;+\; \cdots$$

This and the equation

$$\frac{1}{1 + x} = 1 - x + O(x^2)$$

should make everything clearer.

(I just edited this post a lot. I hope nobody has been quoting this post right now...)

Last edited: Sep 13, 2008
6. Sep 13, 2008

### Hurkyl

Staff Emeritus
But what others have you been taught? I suspect that nothing 'tricky' is going on in that answer -- they simply applied one of the various methods for computing residues that you forgot.

If you want to compute the Laurent series about a pole, then yes, attempting a Taylor series expansion wouldn't be useful, because it doesn't exist. Since you presumably should be learning Laurent series, it would be worth reviewing methods one might use to compute them.

Don't forget your algebra! You know how to factor $z^n + 1$ completely (i.e. into linear factors)... because you know all n roots.

7. Sep 13, 2008

### jostpuur

This equation is not right because of the minus sign.

8. Sep 13, 2008

### Lyuokdea

oops, the minus sign is a typo...

you can calculate laurent series via laurent's theorem, which gives you a series of positive and negative power, and then you calculate a_n and b_n via contour integrals around the point, i'm not sure how that's helpful here, because the contour integral is what we're trying to calculate here....

edit: talked to another friend of mine, for rational functions and at a point with a zero of multiplicity 1: f(x) = g(x)/h(x) has residue = g(x)/ h'(x)... that's not in my book for some reason (which also didn't have the problem in question)

~Lyuokdea

9. Sep 13, 2008

### Hurkyl

Staff Emeritus
That's one of the directions I was heading. I had assumed you were using the book that contained the problem!

By the way... just because you got the answer one way doesn't mean you should stop working on the problem! I think each of the other ideas you've had should really be carried out until the end, until you can solve the problem in several ways!

Incidentally, if you suspect f(z) has a pole of order 1 at a... then (z-a)f(z) is nonsingular at a....

10. Sep 13, 2008

### quasar_4

Just remember to use that formula you have to be working with a function where the pole is simple, g(z) has to be analytic and non-zero at the pole, and h(z0) = 0. It is possible to not always be able to use the formula.