# Homework Help: Contour integral with multiple singularities inside domain without residue theorem?

1. Sep 29, 2010

### newmike

Contour integral with multiple singularities inside domain without residue theorem??

1. The problem statement, all variables and given/known data
Evaluate

$$\oint\frac{dz}{z^{2}-1}$$

where C is the circle $$\left|z\right|$$ = 2

2. Relevant equations

Just learned contour integrals, so not much.
Ok to use Cauchy's Integral formula (if applicable)
Cannot use reside theorem as we haven't learned that yet

3. The attempt at a solution
I've tried this a few different ways...

Anyway, it is clear that there are two singularities: z=1 and z=-1, and both of which are inside the contour. Because of this, I don't think I can use Cauchy's Integration formula.

By attempting the integral the long way and writing out partial fractions, I've arrived at:

$$\frac{1}{2}\oint \frac{2ie^{it}dt}{2e^{it}-1}-\frac{1}{2}\oint \frac{2ie^{it}dt}{2e^{it}+1}$$

where each integral is integrated over 0 to 2*pi

which is where I'm stick. I see that I can turn this into (using substitution):

$$\frac{1}{2}ln(\frac{2e^{it}-1}{2e^{it}+1})$$

evaluated on 0 to 2pi which ends up with zero since it's not analytic. It seems I are ignoring the singularities here...

I have a gut feeling the answer shouldn't be zero... Any suggestions?

Thanks,
-Mike

Last edited: Sep 29, 2010
2. Sep 29, 2010

### Dick

Re: Contour integral with multiple singularities inside domain without residue theore

How about splitting 1/(z^2-1) using partial fractions into the sum of two functions which each have only one singularity?

3. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

I think that is what OP already did.

4. Sep 29, 2010

### newmike

Re: Contour integral with multiple singularities inside domain without residue theore

Well I don't understand how that works since the domain still has two singularities. Are you suggesting apply cauchy's integral formula one at a time. Is that valid? I'm new to the topic and trying to get my head around it. I'll try it out to see what I get.

5. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

I believe the problem is that the function should be Log(z) not ln(r)

Log(z) = ln |r| + i arg(z)

6. Sep 29, 2010

### newmike

Re: Contour integral with multiple singularities inside domain without residue theore

Ok, let me see if that changes things. Thanks.

7. Sep 29, 2010

### Dick

Re: Contour integral with multiple singularities inside domain without residue theore

Yes, it does look like the OP already did partial fractions. But my version of the Cauchy integral formula tells my how to integrate 1/(z-1) without using a contour parametrization. I don't see why you need to use one.

8. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

log(z) = ln|z| + iarg(z)

not

log(z) = ln|r| + iarg(z) .

Sorry for the mistake :)

9. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

I edited my post. I didn't read the post completely myself :)

But following your suggestion would also be useful to OP.

EDIT

The question as not adressed to me but yes it is valid.

Last edited: Sep 29, 2010
10. Sep 29, 2010

### newmike

Re: Contour integral with multiple singularities inside domain without residue theore

Ok, I used this and still got zero.

I'm going to try use cauchy's integral formula on each one as you both are suggesting. Thanks again.

11. Sep 29, 2010

### Dick

Re: Contour integral with multiple singularities inside domain without residue theore

If you get zero again, I think you are on the right track.

12. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

Anyway follows Dick suggestion it is a better one. I gave you the suggestion so that you didn't have to rework anything.

The zero is because the sum of the singularities "cancel" out.

13. Sep 29, 2010

### newmike

Re: Contour integral with multiple singularities inside domain without residue theore

Thanks for all the help guys,

I applied cauchy's integral formula on both integrals and I also got zero.

Using: $$\oint\frac{dz}{z-1}$$

I let f(z)=1, and f eval'd at the singularity is obviously f(1)=1, so I said

$$\oint\frac{dz}{z-1}$$ = 2$$\pi$$i(1) = 2$$\pi$$i

Likewise, I got the same answer for the other integral because again f(z)=1 so f(-1)=1.

Altogether:

1/2 * 2 pi i - 1/2 * 2 pi i = 0

Anything noticeably wrong??..or should I just accept the fact that the answer is indeed zero ;) Thanks again!!

14. Sep 29, 2010

### ╔(σ_σ)╝

Re: Contour integral with multiple singularities inside domain without residue theore

Accept it :).

15. Sep 29, 2010

### newmike

Re: Contour integral with multiple singularities inside domain without residue theore

Ok I'll accept it ;)

That makes sense. Now I feel more confident with the result of zero.

Thanks again, both of you, I've gone crazy over this one problem!

Take care,
-Mike