Solve $\int_{|z|=3}\frac{\mathrm{e}^{\frac1{1-z}}}z\,\mathrm{d}z$ I have to solve following integral:

$$\int_{|z|=3}\frac{\mathrm{e}^{1/(1-z)}}z\,\mathrm{d}z$$

So I decided to use the Residue Theorem. This function has two singularities at the points $z=0$ and $z=1$.
$z=0$ is a simple pole and the residue is equal to $\mathrm{e}$ at this point. But $z=1$ is an essential singularity. I tried to write the Laurent series, and I am able to do it for $\mathrm{e}^{1/(1-z)}$, but I don't know what to do about $\frac1z$. Maybe I have to change variable?
Any hints? Thanks!
 A: Note that the only singularities are at $0$ and $1$.  Hence, from Cauchy's Integral Theorem, we have for any $R>3$
$$\begin{align}
\oint_{|z|=3}\frac{e^{\left(\frac1{z-1}\right)}}{z}\,dz&=\oint_{|z|=R}\frac{e^{\left(\frac1{z-1}\right)}}{z}\,dz\\\\
&\overbrace{=}^{z=Re^{i\phi}}i\int_0^{2\pi}e^{\left(\frac{1}{Re^{i\phi}-1}\right)}
\,d\phi\\\\
\end{align}$$
Letting $R\to \infty$, we find that 
$$\oint_{|z|=3}\frac{e^{\left(\frac1{z-1}\right)}}{z}\,dz=2\pi i$$
A: Someone has already posted a separate (and in my opinion quite ingenious) answer but it is also not too difficult to find the first coefficient in the  Laurent expansion at $z = 1$. We see in a punctured neighborhood of $z = 1$, $$\frac{e^{1/(1-z)}}{z} = \frac{e^{1/(1-z)}}{1 - (1-z)} = \left(\sum_{n\ge 0} \frac{(1-z)^{-n}}{n!} \right)\left(\sum_{n\ge 0} (1-z)^n \right).$$ Now using the Cauchy product formula: $$\left(\sum_{n\ge 0} a_n \right)\left(\sum_{n\ge 0}b_n \right) = \sum_{n\ge 0} \sum_{0 \le k \le n} a_k b_{n-k}.$$ We have $$\frac{e^{1/(1-z)}}{z} = \sum_{n\ge 0} \sum_{0 \le k \le n} \frac{(1-z)^{-k}}{k!} (1-z)^{n-k}= \sum_{n\ge 0} \sum_{0 \le k \le n} \frac{(1-z)^{n-2k}}{k!}.$$ Now if $n$ is even, then $n-2k \neq -1$ for any $k$. If $n$ is odd then $n = 2m-1$ for some $m=1,2,\ldots$ and $n-2k = -1$ for $k = m$. This shows that the coefficient of $(1-z)^{-1}$ in this expansion is $$\sum_{m\ge 1} \frac{1}{m!} = e-1,$$ but we need the coefficient of $(z-1)^{-1}$ so we take the negative: $1-e$. Thus since the residue at $z = 0$ is $e$, we have  $$\oint_{\{\lvert z \rvert = 3\}}\frac{e^{1/(1-z)}}{z} \mathrm{d}z= 2\pi i(\text{sum of residues}) = 2\pi i(e + (1-e)) = 2\pi i.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\begin{align}
\oint_{\verts{z} = 3}{\expo{1/\pars{1 - z}} \over z}\,\dd z &
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z} = 1/3}{\expo{z/\pars{z - 1}} \over z}\,\dd z =
\overbrace{\oint_{\verts{z} = 1/3}{\dd z \over z}}^{\ds{=\ 2\pi\ic}}\ +\
\sum_{n = 1}^{\infty}{1 \over n!}\
\overbrace{\oint_{\verts{z} = 1/3}{z^{n - 1} \over \pars{z - 1}^{n}}\,\dd z}
^{\ds{=\ 0}}
\\[5mm] & = \bbx{2\pi\ic}
\end{align}
