Hey guys.

How can I calculate the residues of this function (in the pic) in all of its singularity points?

I'm kind of a newbie in this this residues stuff and I can really use an example.

Thanks in advance.

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- Feb 9th 2009, 09:36 AMasi123residues
Hey guys.

How can I calculate the residues of this function (in the pic) in all of its singularity points?

I'm kind of a newbie in this this residues stuff and I can really use an example.

Thanks in advance. - Feb 9th 2009, 09:53 AMMoo
- Feb 9th 2009, 10:01 AMMoo
Also, be careful of the

*removable singularities*!

Poles are points where the limit is undefined.

But removable singularities are points where the limit exists. For example $\displaystyle \frac{z^2-3z+2}{(z-2)(z-4)}$

2 annulates the denominator, so you can think that it's a pole. But (z-2) also divides zē-3z+2 (because zē-3z+2=(z-1)(z-2))

So the limit when z goes to 2 is defined. It's a removable singularity. - Feb 9th 2009, 10:04 AMasi123
- Feb 9th 2009, 10:13 AMMoo
No you don't, because $\displaystyle z^2+z-1=\left(z-\frac{-1+\sqrt{5}}{2}\right)\left(z-\frac{-1-\sqrt{5}}{2}\right)$

it was just in case...because you might meet some in the future :P

There are also essential singularities, which are singularities that are neither poles nor removable singularities.

http://en.wikipedia.org/wiki/Mathema...mplex_analysis - Feb 10th 2009, 12:45 AMasi123
Ok, this is what I did (in the pic).

Now I need to sum the residues to get the answer?

Thanks. - Feb 11th 2009, 08:44 AMMoo
It depends on where you integrate o.O