Find the residues of the function

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Homework Statement



Find the residues of the function

Homework Equations



f(z)=[itex]\frac{1}{sin(z)}[/itex] at z=0, [itex]\frac{∏}{2}[/itex], ∏

The Attempt at a Solution


Since the function has a simple pole at z=0

I used: Res(f,0)=lim[itex]_{z->0}[/itex](z-0)[itex]\cdot[/itex][itex]\frac{1}{sin(z)}[/itex]=1. This means the residue of the function at z=0 is 1.

And Res(f,[itex]\frac{∏}{2}[/itex])=lim[itex]_{z->∏/2}[/itex](z-[itex]\frac{∏}{2}[/itex])[itex]\cdot[/itex][itex]\frac{1}{sin(z)}[/itex]=0. This means the residue of the function at z=[itex]\frac{∏}{2}[/itex] is 0.

However I think this method can not be applied on solving z=∏. How can I work it out?
 
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Answers and Replies

  • #2
vela
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Try expanding sin z in a series about ##z=\pi##. Or, better yet, use ##\sin z = \sin[(z-\pi)+\pi]##. The basic idea is to get everything in terms of ##z-\pi##.
 
  • #3
Try expanding sin z in a series about ##z=\pi##. Or, better yet, use ##\sin z = \sin[(z-\pi)+\pi]##. The basic idea is to get everything in terms of ##z-\pi##.
If I turn everything into z-∏, do we need to use the same method as I used to work out z=0 and z=[itex]\frac{∏}{2}[/itex]?
But how? Turn the limit to z-∏→0?
Because I used computer to work out the answer of z=∏ which is -1. But I still cannot get this answer by myself.
 
  • #4
vela
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You find the residue by evaluating the limit
$$\lim_{z \to \pi}\ (z-\pi)\frac{1}{\sin z}.$$ Rewriting the limit in terms of ##z-\pi## is simply to make the evaluation easier. I just realized you could simply apply L'Hopital's rule to do that and avoid unnecessary complications. If you had a more complicated function, however, writing things in terms of ##z-\pi## is often less work than using L'Hopital's rule.

But say you did it with using the trig identity anyway. You should end up with
$$\lim_{z\to\pi} \frac{z-\pi}{-\sin(z-\pi)}.$$ You might already recognize that limit, but it not, use the substitution ##w=z-\pi## to turn it into one you should recognize.
 

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