Domain for Laurent Series

Question

I need to evaluate $f(z)=\frac{e^{-z}}{(z-1)^2}$ around the circle $|z|=3$. I am trying to figure out which domain the Laurent series will be valid. Is it $-1<|z-1|<3$?

Or would it be analytic everywhere $0<|z-1|< \infty$? I am not sure if I have to confine it in the circle or not.

Answer 1

Since the only singularity of your function is at $z=1$, there will be a Laurent series about $z=1$ valid for $0 < |z-1| < \infty$.