Singularities in C* of f(z) = \frac{{\pi z - \pi {z^3}}}{{\sin (\pi z)}}

In summary, the singularities in C* of f(z) = \frac{{\pi z - \pi {z^3}}}{{\sin (\pi z)}} are z = n, with n \in Z, n\neq 0, n\neq 1, and z = 0 and z = 1 are removable singularities. The residue at z = 0 is 0, and the residue at infinity is the same as the residue at 0 of f(1/z), which can be found by looking at f(\frac{1}{z}) = \frac{\frac{\pi}{z} - \frac{\pi}{z^3}}{\sin (\frac{\pi}{z
  • #1
libelec
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0

Homework Statement



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

The Attempt at a Solution



I found that the singularities in C are z = n, with n [tex]\in[/tex] Z, n[tex]\neq[/tex] 0, n[tex]\neq[/tex] 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.
 
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  • #2
Could somebody check this out, please?
 
  • #3
Did I write this correctly? Answer please, because I don't know if I'm translating my question properly into English.
 
  • #4
No one? Please, I need to know this urgently.
 
  • #5
A singularity at infinity of f(z) has the same residue as the singularity at 0 of f(1/z).

Look at
[tex]f(\frac{1}{z}) = \frac{\frac{\pi}{z} - \frac{\pi}{z^3}}{\sin (\frac{\pi}{z})}[/tex]
 
  • #6
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?
 

Related to Singularities in C* of f(z) = \frac{{\pi z - \pi {z^3}}}{{\sin (\pi z)}}

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