# Singularity in the infinity

1. Oct 31, 2009

### libelec

1. The problem statement, all variables and given/known data

Find and classify the singularities in C* of $$f(z) = \frac{{\pi z - \pi {z^3}}}{{\sin (\pi z)}}$$, and give information about Res(f, 0) and Res(f, infinity)

3. The attempt at a solution

I found that the singularities in C are z = n, with n $$\in$$ Z, n$$\neq$$ 0, n$$\neq$$ 1. These are simple poles, while z=0 and z=1 are removable singularities (therefore, Res(f, 0)=0).

Now, in C*: what I thought is that, since the poles tend to infinity when n tends to infinity, then there is a non-isolated singularity.

But then I don't know how to calculate Res(f, infinity).

Did I think that the right way or what am I missing?

Thanks.

Last edited: Oct 31, 2009
2. Oct 31, 2009

### libelec

Could somebody check this out, please?

3. Nov 1, 2009

### libelec

Did I write this correctly? Answer please, because I don't know if I'm translating my question properly into English.

4. Nov 1, 2009

### libelec

No one? Please, I need to know this urgently.

5. Nov 2, 2009

### HallsofIvy

A singularity at infinity of f(z) has the same residue as the singularity at 0 of f(1/z).

Look at
$$f(\frac{1}{z}) = \frac{\frac{\pi}{z} - \frac{\pi}{z^3}}{\sin (\frac{\pi}{z})}$$

6. Nov 2, 2009

### libelec

Thanks. You say the only way to find out is to find the a-1 coefficient (and then multiply by -1)?

Was my assumption that it was a non-isolated singularity correct?