How to calculate residium at infinity?

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SUMMARY

The discussion focuses on calculating the residue of the function \( z^3 \sin \frac{1}{3} \) at infinity. The correct approach involves using the conformal transformation \( u = \frac{1}{z} \) to analyze the behavior of the function as \( z \) approaches infinity. The residue at infinity is defined as the residue at zero of the transformed function \( -\frac{1}{u^2} f\left(\frac{1}{u}\right) \). This method ensures accurate calculation of residues for functions with poles at infinity.

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i have such a function
[tex] z^3 \sin \frac{1}{3}[/tex]

i need to calculate its residium at z=infinity

if i substitue infinity instead of a"" into the formal formula
[tex]res(f(x),a)=\lim_{x->a}(f(x)(x-a))[/tex]

i get infinity
am i correct?
 
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Consider first a pole at some point z = p. You know that 2 pi i times the residue there is the value of the contour integral that encirles that pole anti-clockwise and no other poles. If you then perform the conformal transform u = 1/z in that contour integral, you get the contour integral of:

-1/u^2 f(1/u)

which encircles the corresponding pole in the u-plane at u = 1/p. Note that the contour integral is traversed anti-clockwise if the contour does not encircle the origin. So, the residue at z = p of f(z) is the same as the residue of -1/u^2 f(1/u) at u = 1/p. The residue at infinity is defined such that this result holds in the limit p to infinity. So, it defined as the residue at zero of -1/z^2 f(1/z).
 

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