How do I find the Laurent series of the exponent of $-z$ around e.g. $z=-1$, in the annulus $|z+1|>0$?
Would it just be the Taylor series of the exponent of $-z$ around $-1$ since it converges for all values of $z$ in the annulus?
The Laurent series I'm after is that of $f(z) = \frac{e^{-z}}{(z+1)^2}$ in $|z+1|>0$, and I suppose that if the above way of finding the Laurent series of $e^{-z}$ in this annulus is correct I can divide the series by ${(z+1)^2}$ to find the desired series of $f(z)$.
The key point is that $e^{-z}$ is analytic not only in the annulus, but also at $z = -1$. That is, it is analytic in the disc $|z+1| \geq 0$. For this reason, it has a Taylor expansion at $z = -1$, which is also its Laurent expansion. It then suffices to divide this expansion by ${(z+1)}^2$, which yields a Laurent expansion for $f(z)$.