Residue at infinity, contradiction?

Question

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?

Answer 1

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 $a_{-1}$.

Answer 2

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.