Evaluate : $\oint_C (1+z+z^2)[e^{2/z}+e^{2/(z-2)}+e^{2/(z-3)}]dz$ over the circle $C:|z|=\frac{1}{4}$.
It is clear that terms that involve $e^{2/(z-2)}+e^{2/(z-3)}$ need not to be considered here as they have singularities outside the given region. But I have problem to calculate the residue (at the pole $0$) within the region of the other remaining terms here. Then we should apply Cauchy's residue theorem to evaluate that integral. So how to calculate the residue at the pole $0$?
On
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\oint_{\verts{z} = 1/4}\pars{1 + z + z^{2}} \left[\expo{2/z} + \expo{2/\pars{z - 2}} + \expo{2/\pars{z - 3}}\right]\dd z = \oint_{\verts{z} = 1/4}\pars{1 + z + z^{2}}\expo{2/z}\dd z \\[5mm] \stackrel{z\ \mapsto\ 1/z}{=} &\ \oint_{\verts{z} = 4}\pars{{1 \over z^{2}} + {1 \over z^{3}} + {1 \over z^{4}}}\expo{2z}\dd z = 2\pi\ic\pars{2 + 2 + {4 \over 3}} = \bbx{{32\pi \over 3}\,\ic} \end{align}
HINT:
$$e^{2/z}=\sum_{n=0}^\infty \frac{2^n\,z^{-n}}{n!}=1+\frac2z+\frac2{z^2}+\frac{4}{3z^3}+\cdots $$
Now, multiply $e^{2/z}$ by $1$, $z$, and $z^2$ and determine the residues from each term separately.