<s>[INST] Residue at essential singularity I need a little help with the following problem. I've tried many ways, but i didnt succeed. I think there needs to be a trick or something, some transformation. The task is to find the residue of the function at its singularity e.g. z=-3
\begin{equation}
f(z)=\cos\left(\frac{z^2+4z-1}{z+3}\right)
\end{equation}
I tried to write it as
\begin{align}
\cos\left(\frac{z^2+4z-1}{z+3}\right)=1-\frac{1}{2!}\left((z+1)-\frac{4}{z+3}\right)^2+\frac{1}{4!}\left((z+1)-\frac{4}{z+3}\right)^4-\frac{1}{6!}\left((z+1)-\frac{4}{z+3}\right)^6+\ldots
\end{align}
and collect the coefficients at $\frac{1}{z+3}$ using binomial expansion of the brackets, but it seems to be a dead end, because there is to much of them and well hidden. If somebody could give me a hint, that would be great. Thanks.
 [/INST] $$
\begin{align}
&\cos\left(\frac{z^2+4z-1}{z+3}\right)\\
&=\cos\left((z+3)-2-\frac4{z+3}\right)\\
&=\cos\left((z+3)-\frac4{z+3}\right)\cos(2)+\sin\left((z+3)-\frac4{z+3}\right)\sin(2)\tag{1}
\end{align}
$$
Since $\cos\left((z+3)-\frac4{z+3}\right)$ is an even function of $z+3$, its residue at $z=-3$ is $0$.
The $(z+3)^{-1}$ term of
$$
\frac{(-1)^n}{(2n+1)!}\left((z+3)-\frac4{z+3}\right)^{2n+1}\tag{2}
$$
is
$$
\begin{align}
&\frac{(-1)^n}{(2n+1)!}\binom{2n+1}{n}(z+3)^n\left(-\frac4{z+3}\right)^{n+1}\\[6pt]
&=\frac{-1}{(2n+1)!}\binom{2n+1}{n}\frac{4^{n+1}}{z+3}\tag{3}
\end{align}
$$
Summing and multiplying by $\sin(2)$, we get the residue to be
$$
\begin{align}
-\sin(2)\sum_{n=0}^\infty\frac{4^{n+1}}{n!(n+1)!}
&=-2\sin(2)I_1(4)\\
&=-17.7485131\tag{4}
\end{align}
$$
where $I_n(z)$ is the modified Bessel function of the first kind.
</s>