Laurent series $e^{\frac{1}{1-z}}$ at $|z| > 1$

Question

Firstly, I use this: $e^{\frac{1}{1-z}} = \sum_{n=0}^{\infty} \frac{(-1)^n \frac{1}{z}}{n!\big(1 - \frac{1}{z}\big)^n}$ Next I use the binomial expansion for $\big(1 - \frac{1}{z}\big)^n = \sum_{k=0}^{n} \frac{n! (-1)^k z^{-k}}{k!(n-k)!}$ Combining this, I have: $e^{\frac{1}{1-z}} = \sum_{n=0}^{\infty} \frac{(-1)^n \frac{1}{z}}{n! \sum_{k=0}^{n} \frac{n! (-1)^k z^{-k}}{k!(n-k)!}}$ How I can simplify it? Or maybe there is another way to get a Laurent series? Thank you in advance! Nothing on stack help me.

Answer 1

Write $\exp{\big(\frac{1}{1-z}\big)}=\exp{\big(\frac{1/z}{1/z-1}\big)}.$ Why? Because we're going to use the generating function for the Laguerre polynomials, $$(1-y)^{-a-1}\exp{\big(\frac{x\,y}{y-1}\big)} = \sum_{n=0}^\infty L_n^{a}(x)\,y^n \quad \text{where} \quad L_n^{a}(x) = \sum_{m=0}^n\frac{(-x)^m}{m!}\binom{n+a}{n-m} $$ Set $y=1/z, \, a=-1, \, \text{and } x=1$ and the problem is finished.