Let $a \in \mathbb C$ be a complex number, and consider the function $$ f(z) = \frac{e^{1/z}}{z+a}$$ Compute the residues of $f(z)$ at each of its singularities.
So I know there are singularities at $z=0$ and $z=-a$, and I know $\text{Res}(f(z), z_0) = g(z_0) / h'(z_0)$ can be used for the singularity at $z=-a$ to find the residue to be $e^{-1/a}$, but am not sure how to compute the residue at $z=0$. Any help would be appreciated.
$$\begin{align}e^{1/z}&=\sum_{n=0} \frac{1}{n!}\cdot \frac{1}{z^n}\\ \\ \frac{1}{z+a}&=\frac{1}a\cdot\frac{1}{1-\left(-\frac{z}a\right)}=\sum_{n=0}\frac{(-1)^n}{a^{n+1}}z^n \end{align}$$
Now, we need to collect the coefficients for the term $1/z$,
$$Res=\sum_{n=0} \frac{1}{(n+1)!}\cdot \frac{(-1)^n}{a^{n+1}}=1-\sum_{n=0}\frac{1}{n!}\cdot \left(\frac{-1}{a}\right)^n=1-e^{-\frac{1}a}$$