# Laurent Expansion

1. Apr 10, 2009

### latentcorpse

Say you have $f(z)=\frac{1}{(z+i)^2(z-i)^2}$

a past exam question asked me to find and classify the residues of this.
i had to factorise it into this form and then i just said there was a double pole at $z=+i,z=-i$

now for 5 marks, this doesn't seem like very much work.

is it possible to perform a laurent expansion and then show explicitly that they are poles of order 2 rather than just saying "the power of the brackets is 2 and so it must be of order 2"?

2. Apr 10, 2009

### Dick

If you want to show the explicit expansions you'll need to do two of them. One around the pole at z=i and other around the pole at z=(-i).

3. Apr 10, 2009

### Hurkyl

Staff Emeritus
I can't fathom why you would want to compute the Laurent series to find the singularities; just use what you know about poles and arithmetic. Making the problem harder simply for the sake of making it harder is not what mathematics is about....

I note, however, that the question you stated is to find the residues, and you haven't done anything on that....

4. Apr 10, 2009

### latentcorpse

so you mean do them seperately?

also how would i expand these? using $(1+z)^n=1+nz+\frac{n(n-1)}{2!}z^2+...$? (what the heck is this expansion called anyway - it's not binomial is it?)

5. Apr 10, 2009

### latentcorpse

yeah ive actually done the entire question. im just wondering if i need to say more about the classification of the poles to get all the marks in the exam or is what i put in post 1 enough?

6. Apr 10, 2009

### Dick

Yes, it's enough. E.g. 1/(z-i)^2 is a double pole around z=i and 1/(z+i)^2 is analytic in the neighborhood of z=i. A function doesn't have a single laurent series. It has a different laurent series around every point z=a in the complex plane.