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

Question

I do this like $\dfrac{e^z}{1-z}$ = $-e \dfrac{e^{z-1}}{z-1}$ = $-e \sum_{k=0}^{\infty} \dfrac{(z-1)^k}{k!(z-1)}$ = $-e \sum_{k=0}^{\infty} \dfrac{(z-1)^{k-1}}{k!}$

However doesn't this give me the Laurent series around $|z-1| > 0$? Should I do a taylor expansion of the first term $\dfrac{-e}{z-1}$ to a geometric series to get the expansion for $|z|>1$?

Maybe I have just messed up some definitions, sorry if that's the case.

Answer 1

$|z|>1$ is an annulus with center $0$, thus all powers of the relevant Laurent series should be of the form $(z-0)^k$.

On the one hand $$e^z=1+z+\frac{z^2}{2}+ \dots=\sum_{k=0}^\infty \frac{z^k}{k!}$$

On the other $$\frac{1}{1-z}= -\frac{1}{z} \frac{1}{1-1/z}=-\frac{1}{z} \sum_{k=0}^\infty \frac{1}{z^k}=\sum_{k=0}^\infty- \frac{1}{z^{k+1}} $$

Try multiplying both series and collecting the terms of the same degree.

Answer 2

Put $1 - z = u$. So you shall get $\frac{e^z}{1 - z} = \frac{e^{1-u}}{u} = e \frac{e^{-u}}{u}$.

Now expand it in terms of $e^{-u}$ and get $e \left( \frac{1}{u} - 1 + \frac{u}{2!} - \dots \right)$

Now put the value of $U$ and finally you shall get

$$\frac{e^{z}}{1 - z} = e (\frac{1}{1-z} - 1 + \frac{1 - z}{2!} \dots )$$

The function has a pole of order $1$ at $z = 1$.