What is the contour integral of multiple poles... I'm struggeling to integrate the following where the contour $|z|=1$ because the residues seem to include 'z':
$$
\oint_C \frac{e^{1/z}}{z-a} dz
$$
To find the residues, I first put the function into a power series:
$$
\frac{e^{1/z}}{z-a} = \frac{1}{z-a} + \frac{1}{z(z-a)} + \frac{1}{2!z^2(z-a)}+ \frac{1}{3!z^3(z-a)}+...
$$
From here we see a simple pole at $z_0=a$ which implies the first residue $b_1=e^{1/a}$
and that the solution might be $2\pi i e^{1/a}$ ?
I'm really not sure as there seems to be another pole at $z_0=0$...
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
& \color{#44f}{\left.\oint_{\verts{z}\ =\ 1}\,\,\,{\expo{1/z} \over
z - a}\,\dd z\right\vert_{\verts{a}\ <\ 1}}
\sr{z\ \mapsto\ 1/z}{=}
\oint_{\verts{z}\ =\ 1}\,\,\,{\expo{z} \over
1/z - a}\,{\dd z \over z^{2}}
\\[5mm] = & \
-\,{1 \over a}\oint_{\verts{z}\ =\ 1}\,\,\,{\expo{z} \over
z\pars{z - 1/a}}\,\dd z =
-\,{1 \over a}\braces{2\pi\ic\,\on{Res}\bracks{{\expo{z} \over
z\pars{z - 1/a}},z = 0}}
\\[5mm] = & \
-\,{1 \over a}\bracks{2\pi\ic\,{1 \over \pars{-1/a}}} = \bbx{\color{#44f}{2\pi\ic}} \\ &
\end{align}
