Problems with residue at infinity I know I've asked a similar question before, but this function gives me headaches.
Let $f(z)=\frac{e^z}{z^2*(z^2+9)}.$ Calculate residue in isolated singularity (inclusive infinity)
I calculated the residue in all singularities: $3i,$ $ -3i$ and $ 0, $ but at infinity I got stuck.
 A: Method 1:  Consider the circle of radius $4$ with positive orientation.  On one side, the three poles you mention in your question.  On the other side, all the other poles.  Since the denominator only gives the three poles you mention and the numerator only gives a pole at infinity, and the function is meromorphic, putting the negative orientation on the circle, it only encloses the pole at $\infty$.  Since reversing the path of an integral negates its value, the residue of pole at $\infty$ is minus the sum of the residues of the poles you found.
Method 2:  For a residue at infinity, let $z \mapsto 1/w$ and expand $(-1/w^2)f(1/w)$ in a Laurent expansion in the variable $w$.  (Recall that a residue is an integral.  The change of variable changes the differential element: $\mathrm{d}z = \frac{\mathrm{d}}{\mathrm{d}w} w^{-1} \,\mathrm{d}w = -w^{-2}\,\mathrm{d}w$, which explains the source of the $(-1/w^2)$.)  Here, this gives \begin{align*}
\frac{-1}{w^2}f(1/w) &= \frac{-w^2 \mathrm{e}^{1/w}}{1+9w^2}  \\
&= -w^2 \left(\sum_{j=0}^\infty \frac{(1/w)^j}{j!}  \right) \sum_{k=0}^\infty (-9w^2)^k  \text{.}\end{align*}
To get the coefficient of the $w^{-1}$, we need the coefficient of $w^{-3}$ in $-\left(\sum_{j=0}^\infty \frac{1}{j!}w^{-j}  \right) \sum_{k=0}^\infty (-9)^kw^{2k}$.  Contributions to that coefficient occur when $2k-j = -3$, so, $j = 2k+3$.  ... blah, shift the index, blah, recognize a sine series (maybe), blah ...  This seems hard because the powers are running in opposite directions (through negative integers in $j$ and through positive evens in $k$).
Method 2 (fixed): As in the previous attempt, but make the powers run in the same direction.
\begin{align*}
\frac{-1}{w^2}f(1/w) &= \frac{-w^2 \mathrm{e}^{1/w}}{1+9w^2}  \cdot \frac{1/(9w^2)}{1/(9w^2)}  \\
&= \frac{-1}{9} \frac{\mathrm{e}^{1/w}}{(1/9)w^{-2} + 1}  \\
&= \frac{-1}{9} \left(\sum_{j=0}^\infty \frac{w^{-j}}{j!}  \right) \sum_{k=0}^\infty \left(\frac{-1}{9} w^{-2}\right)^k  \text{.}
\end{align*}
(Another way to say this:  If you are expanding in powers of $1/w = w^{-1}$ make sure you expand everything in powers of $w^{-1}$.  Your change of variable has moved the singularity to $0$ so to make sure you are working with series convergent on the annulus with inner radius $0$, choose the power series that blow up there.)  Now the $w^{-1}$ term receives contributions when $-j-2k=-1$, $j+2k=1$, so only from the $j = 1$ and $k = 0$ terms.
$$  \frac{-1}{9}\left(\frac{w^{-1}}{1!}\right)\left(\frac{-1}{9} w^{-2}\right)^0 = \frac{-1}{9}  $$
and we get the same residue as from Method 1.
Could we avoid using a big theorem (Method 1) or messing around with series (Methods 2)?  Not really; the singularity at $\infty$ of $\mathrm{e}^z$ is at infinity and is essential, so the usual tricks ($\lim_{z \rightarrow \infty} (z-\infty)f(z)$ for a simple pole and $\frac{1}{(n-1)!} \lim_{z \rightarrow c} \frac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}} ((z-\infty)^n f(z))$) are useless.  This means either we find a theorem to do the work for us, or we have to use Laurent expansion.
