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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Is it possible you are using the wrong poles? That might make the answer change sign.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I don't think so... type this into Wolfram Alpha:

"residues of the function sqrt(z)/(1+z^3)"

It gives i/3, i/3 and -i/3, the sum of which is i/3.

2iπ(i/3) = -2π/3... but I don't know why.....

**zetafunc.****Guest**

Never mind, I figured it out. It's not me that's wrong, it's Wolfram. The arguments of the poles should all be positive, but for some reason Mathematica doesn't do that.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Those are the correct residues.

I have a worked example using a keyhole contour. It is slightly different than your problem but it might help to clear up what you are doing wrong.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I think those residues I gave are wrong. If you use positive arguments/angles for the poles (i.e. 5pi/6 instead of -pi/3) you will get the correct residue. The reason for this discrepancy is that Mathematica defines the negative real axis as the branch cut, when it is actually the positive axis.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Okay, glad it worked out.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

The lesson here is that you should never be 100% certain about a CAS's answer... right?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

The way I understand it and not too well at that is that M has to make a decision about the branch cuts from a number of choices. They set it up to pick the most useful one. But that may not always be the one you want.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

So it is clearly best to compute the residues for these types of problems by hand, then.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

When possible, you should always be checking their results...

For simple poles the residues will not be difficult.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

What I mean is that it is probably best not to tell a CAS to find the residues of f(z). Instead, input the calculation for the residues, so there is no ambiguity. (Or at least, far less.)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

That is how I get mine.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

So what did you need help understanding in my keyhole contour of that problem? I did skip over some steps, I can explain them if it would help.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Hi;

Anything that you would like to add to the original post would be fine for me as well as anyone else.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I cannot edit posts but I can post it again. What should be added?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

HI;

you do not see this at the bottom of your posts?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

No, I only see the Quote button.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Do you see my avatar?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

I see, it is because you are a guest.

For the keyhole contour you do not have to add anything. It is fine. But if you want to post more about it you can of course.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I think there might be some keyhole contour problems in the Schaum's book. There is definitely an example in there that uses exactly the same method that I used in that post.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Sorry, I was having back problems and had to get on the floor for a couple of hours.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

What happened?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,362

Nothing yet, I feel pretty bad but I am hungry so I have to get up and sit like a human being.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

You were on the floor for hours? That sounds pretty serious.