Integral of (1/8z^3 -1) around Contour C=1: Step by Step Guide

[soulsearching]

Also when trying to find the integral of (1/8z^3 -1) around the contour c=1.

I found the singularities to be 1/2, 1/2exp(2pi/3), and 1/2exp(4pi/3)

What is the next step here. Do I just assume the integral is 6pi(i) after using partial fractions to find the numerators of the 3 fractions.

The answer is actually 0, but I don't understand how it was solved. I know it might have something to do with the rule that says, the sum of the integral of a C1, C2 and C3 is equal to the integral of the outside contour, but how do i know the orientation of the three contours inside the big contour C.

Thanks, hope I am not asking too many questions

 

[Pere Callahan]

Find the function's residue at each of the poles inside the contour. Then use the residue theorem.

 

[soulsearching]

Find the function's residue at each of the poles inside the contour. Then use the residue theorem.

We haven't covered residue theorem yet. Is there any other way to solve it?

 

[Pere Callahan]

We haven't covered residue theorem yet. Is there any other way to solve it?

Of course there is. Find a parametrization of the contour and do it by hand.

In general if you want to evaluate
$$\int_C dz f(z)$$

where C is the contour, you can find a parametrization

$$\gamma : [0,1]\to \mathbb{C}$$
such that $\gamma([0,1])=C$ (formalities omitted - you are certainly familiar with that )

Then your integral becomes

$$\int_0^1 dt f(\gamma(t))|\gamma'(t)|$$

This will give a not-too eays but doable integral. Note however, that normally it works the other way around. If you have some definite real integral, you can sometimes try to interpret it as a complex integral over a closed contour and use the residue theorem to evaluate it, but that's a different story.

 

[tiny-tim]

What is the next step here. Do I just assume the integral is 6pi(i) after using partial fractions to find the numerators of the 3 fractions.

The answer is actually 0, but I don't understand how it was solved. I know it might have something to do with the rule that says, the sum of the integral of a C1, C2 and C3 is equal to the integral of the outside contour, but how do i know the orientation of the three contours inside the big contour C.

Hi soulsearching!

I'm a little confused (especially by the 6πi).

Are you saying that you know how to integrate (1/2z - 1) round a contour containing z = 1/2?

If so, just use the same contour for all three partition fractions!

 

[soulsearching]

Hi soulsearching!

I'm a little confused (especially by the 6πi).

Are you saying that you know how to integrate (1/2z - 1) round a contour containing z = 1/2?

If so, just use the same contour for all three partition fractions!

Like after breaking up the fraction into 3 fractions using partial fractions, what's the next step?

Thanks... :)

 

[tiny-tim]

Life after breaking up the fraction into 3 fractions using partial fractions, what's the next step?

erm … if you don't know how to integrate (1/2z - 1) round a contour containing z = 1/2, then there is no next step … you're stuck (or, rather, you have to follow Pere Callahan's method)!

So … do you know ? … you didn't say.

 

[soulsearching]

erm … if you don't know how to integrate (1/2z - 1) round a contour containing z = 1/2, then there is no next step … you're stuck (or, rather, you have to follow Pere Callahan's method)!

So … do you know ? … you didn't say.

I think its going to be 2pi(i) because the singularity (1/2) is inside the contour. is this right?

 

[soulsearching]

i did not get Pere Callahan's method. and why did you choose the contour as z=1/2? the question says integrate it around z=1

 

[tiny-tim]

i did not get Pere Callahan's method. and why did you choose the contour as z=1/2? the question says integrate it around z=1

No, I said a contour containing z = 1/2.

You're getting confused between z = 1/2 (a point) and |z| = 1/2 (a circle) … I did warn you about that in another thread … you must write |z| when you mean it, or you'll lose track.

The integral, of course, is the same for any contour that includes the same singularities.

So you could, for example, choose the contour |z| = 100!

And then integrate all three partial fractions round that.

I think its going to be 2pi(i) because the singularity (1/2) is inside the contour. is this right?

erm …

I can't remember! …​

but I know it's 1/2 times something.