Trying to evaluate a complex integral? I have a question where I must evaluate the following integral over a circle where $\lvert z \rvert = 2$. 
$$I = \oint \frac{z^3e^{\frac{1}{z}}}{1+z}dz$$
I have tried the $z^3 = 8 \cdot \mathrm{e}^{3it}$ approach and ended up with:
$$I = i\int_0^{2\pi} \frac{16e^{4it}e^{\frac{1}{2e^{it}}}}{1+2e^{it}}dt$$
But I have no idea what to do from here! Or if this is even the correct approach. Thanks in advance to anyone who knows
 A: This looks ready-made for the residue theorem.  I will assume you know it and go from there.  The pole at $z=-1$ is simple, and its residue is straightforward:
$$\operatorname*{Res}_{z=-1} \frac{z^3 e^{1/z}}{1+z} = -\frac1{e}$$
The pole at $z=0$, however, is essential.  Still, we seek to find the coefficient of $1/z$ in the Laurent expansion of the integrand about $z=0$:
$$\begin{align}\frac{z^3 e^{1/z}}{1+z} &= [z^{-1}] \left [z^3 (1-z+z^2-z^3+\cdots) \left (1+\frac1{z}+\frac1{2! z^2} +\frac1{3! z^3}+\cdots \right )\right ]\\ &= \frac1{4!} - \frac1{5!}+\frac1{6!}-\cdots\\&= \frac1{e}-\frac1{2!}+\frac1{3!}\end{align}$$
The integral is, by the residue theorem,
$$i 2 \pi \left ( -\frac1{e} + \frac1{e}-\frac1{2!}+\frac1{3!}\right ) = -i \frac{2 \pi}{3}$$
A: Since $ \displaystyle f(z) = \frac{z^{3} e^{1/z}}{1+z}  $ is meromorphic outside of the contour,
$$  \displaystyle\int_{|z|=2} \frac{z^{3} e^{1/z}}{1+z}  = -2 \pi i \text{Res} [f(z), \infty]$$
Using the transformation to find the residue at infinity,
$$ \text{Res}[f(z), \infty] = -\text{Res} \Big[\frac{1}{z^{2}} f \left(\frac{1}{z} \right),0 \Big] = - \text{Res} \Big[ \frac{e^{z}}{z^{4}(z+1)} ,0 \Big]$$
$$ = - \frac{1}{3!} \lim_{z \to 0} \frac{d}{dz^{3}}\frac{e^{z}}{z+1} = - \frac{1}{6}\lim_{z \to 0} \frac{e^{z}(z^{3}+3z-2)}{(z+1)^{4}} = \frac{1}{3}$$ 
So
$$ \displaystyle \int_{|z|=2} \frac{z^{3} e^{1/z}}{1+z}  = -2 \pi i \left(\frac{1}{3} \right) = -\frac{2 \pi i}{3}$$
A: The "best" way to compute this integral is certainly via the residue theorem, as Ron and Random Variable did. 
On can also avoid the residue theorem, as follows. 
Since $e^{1/z}=\sum_0^\infty \frac{z^{-k}}{k!}$, one can write 
$$I=\sum_{k=0}^\infty \frac1{k!}\,\int_{\vert z\vert=2} \frac{z^{3-k}}{1+z}\, dz:=\sum_{k=0}^\infty \frac1{k!}\, J_k\, . $$
Next, for each fixed $k$ we have
\begin{eqnarray}J_k&=&\int_{\vert z\vert=2}\frac{z^{2-k}}{1+\frac1z}\, dz\\
&=&\int_{\vert z\vert=2}z^{2-k}\sum_{n=0}^\infty \frac{(-1)^n}{z^n}\, dz\\
&=&\sum_{n=0}^\infty (-1)^n\int_{\vert z\vert=2}\frac{dz}{z^{n+k-2}}\cdot
\end{eqnarray}
Now, the integral $\int_{\vert z\vert=2}\frac{dz}{z^{n+k-2}}$ is equal to $0$ if $n+k-2\neq 1$, i.e. $n\neq 3-k$, and it is equal to $2i\pi$ if $n=3-k$. However, $n=3-k$ never happens if $k>3$ (many thanks to Ron for finding the mistake in a previous version of this post!!). So we have
$$J_k=(-1)^{3-k}\times 2i\pi=-2i\pi\times (-1)^{k}\quad {\rm if}\quad k\leq 3,$$
and $J_k=0$ for all $k>3$. 
It follows that 
$$I=-2i\pi\sum_{k=0}^3\frac{(-1)^k}{k!}=-{2i\pi}\left(1-1+\frac12-\frac16\right)=-\frac{2i\pi}3\cdot $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\on}[1]{\operatorname{#1}}
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 \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}
I & \equiv \bbox[5px,#ffd]{\oint_{\verts{z}\ =\ 2}\, {z^{3}\expo{1/z} \over 1 + z}\,\dd z}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z}\ =\ 1/2}\,\,\,
{\expo{z} \over z^{4}\pars{1 + z}}\,\dd z
\\[5mm] = &
2\pi\ic\,{1 \over 3!}\,\lim_{z \to 0}\,\totald[3]{}{z}
\pars{\expo{z} \over 1 + z} =
\bbx{-\,{2\pi \over 3}\,\ic} \\ &
\end{align}
