What is the Laurent series of $ \exp \! \bigl( - \frac{1}{z} \bigr) $?

Question

I’m thinking that I could simply let $ x = - \dfrac{1}{z} $ in the Maclaurin series for $ e^{x} $: $$ 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots = 1 - \frac{1}{z} + \frac{1}{2! z^{2}} - \frac{1}{3! z^{3}} + \cdots. $$ Is that right?

Answer 1

The Taylor series for $e^x$ is $$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots $$ Now as you have said, let $x = -\frac{1}{z}$. Then we have $$ e^{-1/z} = 1 - \frac{1}{z} + \frac{1}{2z^2} - \frac{1}{3!z^3} + \cdots = \sum_{n=0}^{\infty}\frac{1}{(-z)^nn!} = \sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!} $$ where the Laurent series is $$ \sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!} $$ since to be a Laurent series we need a principal part $\sum_{n=-\infty}^{-1}$ and an analytic part $\sum_{n=0}^{\infty}$