# residues

• Feb 9th 2009, 09:36 AM
asi123
residues
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.

• Feb 9th 2009, 09:53 AM
Moo
Hello,
Quote:

Originally Posted by asi123
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.

z=0 is a double pole.
z=1 is a simple pole.

The general formula for residues in poles of order n is :

$\displaystyle \text{Res}_{z=a} f(z)=\frac{1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \left\{(z-a)^n f(z)\right\}$
• Feb 9th 2009, 10:01 AM
Moo
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 AM
asi123
Quote:

Originally Posted by Moo
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.

But I don't have any removable singularities in my function, right?

Thanks.
• Feb 9th 2009, 10:13 AM
Moo
Quote:

Originally Posted by asi123
But I don't have any removable singularities in my function, right?

Thanks.

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 AM
asi123
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 AM
Moo
It depends on where you integrate o.O