How to calculate residium at infinity?

[nhrock3]

i have such a function
<br /> z^3 \sin \frac{1}{3}<br />

i need to calculate its residium at z=infinity

if i substitue infinity instead of a"" into the formal formula
res(f(x),a)=\lim_{x-&gt;a}(f(x)(x-a))

i get infinity
am i correct?

 

[Count Iblis]

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).