# Integral where I need to classify the singularities

1. May 7, 2007

### 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 $$z = \pm 1$$.

$$\frac{{\sin \left( {\pi z} \right)}}{{z^4 - 1}} = \frac{{\sin \left( {\pi z} \right)}}{{\left( {z - 1} \right)\left( {z + 1} \right)\left( {z^2 + 1} \right)}}$$

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.

2. May 7, 2007

### HallsofIvy

Staff Emeritus
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?

3. May 7, 2007

### AKG

How did you figure out what the singularities are?

4. May 8, 2007

### 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.

5. May 8, 2007

### HallsofIvy

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

6. May 10, 2007

### 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.