Residues for a Complex Analysis Noob

[Master J]

I need to calculate the residue of

( 1 - cos wt ) / w^2

This has a pole of second order at w=0, am I correct?

Now may math book says that a second order residue is given by

limit z goes to z_0 of {[(z-z_0)^2. f(z)]'} where z_0 is the pole

I'm quite new to complex analysis. Could someone perhaps show me how this relates to the above, and how I get the residue from this?

Cheers!

 

[Feldoh]

The residue is defined as being 'b' in the b/(w-a) term in a Laurent series so to find the residue you simply look for functions that look like b/(w-a) or in your case b/w, where b is just some constant that is the residue.

You could computer the residue using your formula, however that appears to be ugly and complicated.

However there is an easier way to find that residue. You're only looking for the polynomial term with a power of -1. Now if we expand out you're function we get:

$$\frac{1}{w^2} - \frac{cos(wt)}{w^2}$$

1/w^2 is all ready in polynomial form. There is NO 1/w dependence in this part, so you can conclude that 1 is not a residue. What about cos(wt)? Well we can write that as an infinite series of polynomial terms and then divide by w^2 and see what the coefficient of the 1/w term is. Right?

 

[Citan Uzuki]

(1-cos (wt))/w^2 has a removable singularity at w=0, not a pole.

 

[Master J]

Are a pole and a singularity no the same thing?

So the term of power -1 is -t/w, so -t is the residue??


Then, if the integral is 2. pi. i SUM Res( f(z) )

my answer is -2. pi. i t ?

 

[Feldoh]

If I expand it out, there is no 1/w term...??

Are a pole and a singularity not the same thing?

There's no 1/w term? So what is the coefficient?

I believe a removable singularity can be expanded around the undefined point such that the function can be defined as continuous. Anyways if you can prove that your singularity is removable you can prove that the residue is 0.

 

[Citan Uzuki]

Are a pole and a singularity no the same thing?

A pole is one type of singularity, a removable singularity is a different type. As Feldoh says, a removable singularity is one where the function can be defined at the singularity in such a way that it remains continuous (and indeed, analytic).

I'm not sure how you're getting -t/w in the expansion. In an earlier version of your post (the one quoted by Feldoh), you apparently wrote that there is no 1/w term, which is correct. So the coefficient of that term is zero.