Evaluating Contour Integral w/ Multiple Singularities

[newmike]

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

Homework Statement

Evaluate

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

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

Homework 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

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

 

[Dick]

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

 

[╔(σ_σ)╝]

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

I think that is what OP already did.

 

[newmike]

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

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.

 

[╔(σ_σ)╝]

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

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

 

[newmike]

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

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

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

 

[Dick]

I think that is what OP already did.

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.

 

[╔(σ_σ)╝]

Btw that should read

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

not

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

Sorry for the mistake :)

 

[╔(σ_σ)╝]

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.

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



But following your suggestion would also be useful to OP.

EDIT

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.

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

 

[newmike]

Btw that should read

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

not

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

Sorry for the mistake :)

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.

 

[Dick]

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.

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

 

[╔(σ_σ)╝]

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.

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.

 

[newmike]

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!

 

[╔(σ_σ)╝]

Accept it :).

 

[newmike]

Ok I'll accept it ;)

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

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