How to find Laurent series of exp(1/z+z)

Question

My approach was to expand both $e^{1/z}$ and $e^{z}$, and multiply them together, but it seems like a lot of work, is there any shortcut that I can use in solving this problem?

Answer 1

Not so much work. Since: $$ e^{z} = \sum_{n\geq 0} \frac{z^n}{n!},$$ $$ e^{\frac{1}{z}} = \sum_{m\geq 0}\frac{1}{m!z^m}$$ for any $h\in\mathbb{N}$ we have that the coefficient of $x^h$ in their product is given by: $$\sum_{m\geq 0}\frac{1}{(m+h)!m!}=I_h(2)$$ where $I_h$ is a modified Bessel function of the first kind. That gives:

$$ \exp\left(z+\frac{1}{z}\right) = I_0(2)+\sum_{h\geq 1} I_h(2)\left(z^h+\frac{1}{z^h}\right).$$