# 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 $$f(z)=\cos\left(\frac{z^2+4z-1}{z+3}\right)$$

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.

\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.