- #1
Benny
- 584
- 0
Hi I'm just working on an integral where I need to classify the singularities of the integrand [tex]\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}}[/tex].
There are singularities at [tex]z = \pm 1, \pm i[/tex] but the ones I'm interested in are at [tex]z = \pm 1[/tex].
[tex]
\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{\sin \left( {\pi z} \right)}}{{\left( {z - 1} \right)\left( {z + 1} \right)\left( {z^2 + 1} \right)}}
[/tex]
My immediate thought was that z = 1 is a simple pole but if it is I should be able to write [tex]\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{h\left( z \right)}}{{\left( {z - 1} \right)}}[/tex] where h is analytic in a neighbourhood of z = 1 and non-zero at z = 1 but this doesn't seem to be the case here. The answer I have seems to be saying that z = 1 is a simple pole but at the moment I can't see how it can be.
Any input would be great thanks.
There are singularities at [tex]z = \pm 1, \pm i[/tex] but the ones I'm interested in are at [tex]z = \pm 1[/tex].
[tex]
\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{\sin \left( {\pi z} \right)}}{{\left( {z - 1} \right)\left( {z + 1} \right)\left( {z^2 + 1} \right)}}
[/tex]
My immediate thought was that z = 1 is a simple pole but if it is I should be able to write [tex]\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{h\left( z \right)}}{{\left( {z - 1} \right)}}[/tex] where h is analytic in a neighbourhood of z = 1 and non-zero at z = 1 but this doesn't seem to be the case here. The answer I have seems to be saying that z = 1 is a simple pole but at the moment I can't see how it can be.
Any input would be great thanks.