Laurent series of functions such as $e^{1/z}$ around $z_0 \ne 0$

Question

I'm having some trouble trying to expand in Laurent series a certain class of functions around points different . These are functions involving terms such as $e^{1/z}, \cos(\pi/z)$ and so on... Even worse when terms of this type are at the denominator of a fraction.

Just to make an example, suppose I want to expand $f(z)=e^{1/z}$ or $g(z)=\frac{1}{e^{1/z}}$ around $z=1$. What am I supposed to do?

In the first case, I am tempted to proceed in the following way: $$e^{1/z}=\sum_{n=0}^\infty \frac{1}{n!} z^{-n}=\sum_{n=0}^\infty \frac{1}{n!} (z-1+1)^{-n}=...??$$

With Taylor series I would use Netwon's binomial formula, but in this case?

In principle I could make the translation $w=z-1$ and try to expand $e^{\frac{1}{w+1}}$ around $w=0$ but this does not seem any better: $$e^{\frac{1}{w+1}}=\sum_{n=0}^\infty \frac{1}{n!} \left(\frac{1}{1+w}\right)^n = \sum_{n=0}^\infty \frac{1}{n!} \left(\sum_{m=0}^\infty (-1)^m w^m\right)^n =...?? $$

Answer 1

You are not likely to find a closed form for $e^{1/z}$ in powers of $z-1$. But of course you can compute as many terms as you have time for. $$ e^{1/z} = {\rm e}-{\rm e} \left( z-1 \right) +{\frac {3\,{\rm e}}{2}} \left( z- 1 \right) ^{2}-{\frac {13\,{\rm e}}{6}} \left( z-1 \right) ^{3}+{ \frac {73\,{\rm e}}{24}} \left( z-1 \right) ^{4}-{\frac {167\,{\rm e} }{40}} \left( z-1 \right) ^{5}+\dots $$

Answer 2

Applying Cauchy product, we can get $$ \left(\sum _{m=0}^{\infty } (-1)^m w^m\right)^n=\sum _{m=0}^{\infty } (-1)^m w^m \binom{m+n-1}{n-1}. $$ As the final result, we have $$ \sum _{m=0}^{\infty } (-1)^m w^m A_m, $$ where $$ A_m=\sum _{n=0}^{\infty } \frac{1}{n!}\binom{m+n-1}{n-1}=\, _1F_1(m+1;2;1). $$