# Thread: locate and classify singularities then compute residue?

1. ## locate and classify singularities then compute residue?

I'm busy with complex analysis and once again i'm having some trouble. Any help would be appreciated.I apologize in advance for the lengthiness of the question I'm just really uncertain about this and would like to understand where my mistakes are.Thanks.

The question is:

Locate and classify all the singularities of the function [sinz/(z^2-pi^2)]. Then compute the residues at each of these singularities.

The only way I know how to do this is using Laurent expansion. I know that the function has isolated singularities at z=pi and z=-pi. I also know the expansion for sinz is sinz=z-z^3/3!+z^5/5!-z^7/7!+.... now for the singularity z=pi we want all terms expressed in terms of (z-pi) so sin(z-pi+pi)= -sin(z-pi)= -[(z-pi)-(z-pi)^3/3!+(z-pi)^5/5!-....] and finding the laurent series for 1/(z^2-pi^2)=1/(z-pi)(z+pi) with 0<lz-pil<2pi we end up with 1/(z-pi)(z+pi) = 1/2pi(z-pi)-1/(2pi)^2+(z-pi)/(2pi)^3-...… we then multiply the series together looking for terms with (z-pi) in the denominator which requires multiplying the first term of 1/(z-pi)(z+pi) by -[(z-pi)-(z-pi)^3/3!+(z-pi)^5/5!-....] and we see that no term will have (z-pi) in the denominator hence this singularity is a removable singularity and the residue=0. The same reasoning applies to z=-pi and we get the same answer except with (z+pi) replacing (z-pi).
I'd like to know if this is correct and if not where I went wrong? If it is correct I'd like help with the following question:

Give an example of a function with pole of order 5 at (1-2i), an essential singularity at 1, and a removable singularity at 0.Justify all assertions.

for a function with pole of order 5 at (1-2i) we need a function with the form: a/(1-2i)+b/(1-2i^2)+...+e/(1-2i)^5 so can we choose a to e arbitrarily then add together to get a single term with (1-2i)^5 as the denominator? for an essential singularity at 1 we need a funtion with the form: a/(z-1)+b/(z-1)^2+..... with infinite terms so can we just choose exp^[1/(z-1)]? and for a removable singularity at 0 we need a function such that no term has any power of z in the denominator so like (1-cosz)/z^2? and then do you just add the three functions together to get 1 function with all three singularities or what?

Thanks in advance for all help.

2. If you look at post number #4 in this it will show you have to find residues without computing the Laurent series

http://www.mathhelpforum.com/math-he...ms-152737.html

Yes your are correct in 1 that they are both removable singularities.

Your idea is sound for the 2nd part but you don't need all of the lower order poles for the first one

$\displaystyle \frac{1}{[z-(1+2i)]^5}$ with have a pole of order 5.

3. thanks for the link I've read about that method but it does not form part of my text or at least I havn't come across it in my text yet. For the second part I'd just like to clarify the last part. Is adding the seperate functions with the different singularities to get one function with all the singularities correct or should the seperate functions be multiplied to get a single function with all the different singularities? Thanks again.