What is the contour integral of multiple poles...

Question

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 $z_0=0$...

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}