Laurent Expansion $\frac{e^z}{z-1}$

Question

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

$t=z-1\iff t+1=z$

$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}=\sum_{n=0}^\infty \frac{(t+1)^n}{n!}$

$\frac{1}{t}=\frac{1}{1-1+t}$

Is it the way?

Answer 1

Simpler:

Don't forget the radius of convergence of the exponential function if infinite), so you can write

$${\rm e}^z={\rm e\, e}^{z-1}={\rm e}\Bigl(1+(z-1)+\frac{(z-1)^2}{2}+\dots+\frac{(z-1)^n}{n!}+\dotsm\Bigr).$$