Residue theorem and laurent expansion

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SUMMARY

The discussion centers on calculating the residue of the function f(z)=\frac{z^{2}}{\sqrt{z^{2}-1}(z-t)} at infinity using Laurent expansion. The original poster questions the necessity of this method, noting that their teacher's approach yields a residue of -t for g(z)=\frac{\sqrt{z^{2}-1}}{z-t}. They express confusion over the direct evaluation at infinity, which results in 1, indicating that a straightforward evaluation does not yield the correct residue. The consensus is that for functions where the Laurent series does not terminate, using the series expansion is indeed the most effective method.

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elimenohpee
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Homework Statement


I need to calculate the residue of a function at infinity. My teacher does this by expanding the function in a laurent expansion and deduces the value from that. That seems much harder than it needs to be. For example, in the notes he calculates the residue at infinity of:
[tex]g(z)=\frac{\sqrt{z^{2}-1}}{z-t}=...=-t[/tex]

Is there an easier way than resorting to a laurent series? If I let z approach infinity in the function above, I get 1 not -t, so I'm assuming you can't evaluate the residue in that way?

Specifically I need to find the residue at infinity of
[tex]f(z)=\frac{z^{2}}{\sqrt{z^{2}-1}(z-t)}[/tex]
but I'm looking for a method to do so.
 
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If you have nicer functions than that there are a few tricks, but when the Laurent expansion doesn't terminate I don't think that there is an easier way than to use a series expansion.
 

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