Integral where I need to classify the singularities

[Benny]

Hi I'm just working on an integral where I need to classify the singularities of the integrand \frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}}.

There are singularities at z = \pm 1, \pm i but the ones I'm interested in are at z = \pm 1.

<br /> \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)}}<br />

My immediate thought was that z = 1 is a simple pole but if it is I should be able to write \frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{h\left( z \right)}}{{\left( {z - 1} \right)}} 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.

 

[HallsofIvy]

So the problem is that sin \pi z has a zero at z= 1. How could you write sin \pi z as a Taylor's series around z= 1?

 

[AKG]

How did you figure out what the singularities are?

 

[Benny]

I found the singularities by setting z^4 - 1 = 0.

The first term in the series for the sin(pi*z), centred at 1, is zero but the second one (which is something multiplied by (z-1)) isn't. So the singularity should be removable but then it isn't a simple pole. It makes no difference to the integral that I was calculating but it's just something that's puzzling me.

 

[HallsofIvy]

Yes, the singularity at z= 1 is removable, not a pole. What made you think it was a pole?

 

[Benny]

The answer said it was a pole. I guess in future I should use my judgement when looking at the answers.

Thanks for the help.