Residue at infinity, contradiction? I am trying to find the residue at $z=\infty$ of the function
$$f(z)=\frac{z^4}{\sqrt{z^2-1}}$$
This has a Laurent series about $\infty$ of:
$$f(z)=
Z^3+ \frac{1}{2}z+\frac{3}{8}\frac{1}{z}+\frac{5}{16}\frac{1}{z^3}+...$$
Something that can be confirmed using wolfram-alpha. From this I would read of the residue at $\infty$ as been
$$\text{res}(f,\infty)=\frac{3}{8}$$
However, when putting this query through wolfram-alpha it  indicates that the residue is:
$$\text{res}(f,\infty)=-\frac{3}{8}$$
If I am wrong in my outcome for the residual why? and if wolfram-alpha is wrong, what could be causing this error?
 A: To get a branch of this function analytic in a neighbourhood of $\infty$, write it as 
$$ f(z) = \dfrac{z^3}{\sqrt{1-1/z^2}}$$
where we take the principal branch of this square root.  This will agree with
Wolfram's Laurent series.  
The residue of $f(z)$ at $\infty$ is defined as the residue of $g(w) = -w^{-2} f(1/w)$ at $w = 0$.  Thus
$$g(w) = - \dfrac{-1}{w^5 \sqrt{1 - w^2}} = \dfrac{-1}{w^5} - \dfrac{1}{2w^3} - \dfrac{3}{8 w} + \ldots $$  
so the residue is indeed $-3/8$.
More generally, if $f(z)$ has a Laurent series $\sum_n a_n z^n$ in a neighbourhood of $\infty$, the residue at $\infty$ is $- a_{-1}$, not $a_{-1}$.
A: The residue at infinity is $-\frac{3}{8}$, a counterclockwise contour integral along a circle of radius R of the function tends to  $2\pi i \times \frac{3}{8}$ when we let R tend to infinity. Since the function is meromorphic everywhere outside the contour it also equals $-2\pi i$ times the sum of all the residues outside of the contour. You can also say that exterior of a contour is also the interior of the contour, provided you invert the winding number. 
