Can $\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{(z-1)} dz$ be solved? How we can calculate the result of following Integral?
$$\oint_{|z|=2}z^3 \bar {z} e^\frac{1}{z-1} \mathrm{d}z$$
 A: $$\begin{align}
\oint_{|z=2|} z^3\bar z e^{\frac{1}{(z-1)}}dz&=\oint_{|z=2|} 4z^2 e^{\frac{1}{(z-1)}}dz\\
&=8\pi i \text{Res}_{z=1}\left(z^2 e^{\frac{1}{(z-1)}}\right)\\
\end{align}$$
The residue $\text{Res}_{z=1} \left(z^2 e^{\frac{1}{(z-1)}}\right)$ can be found as follows
Note that $z^2=(z-1)^2+2(z-1)+1$ and the Laurent expansion of $e^{\frac{1}{z-1}}$ is 
$$e^{\frac{1}{(z-1)}}=1+\frac{1}{z-1}+\frac{1}{2!}\frac{1}{(z-1)^2}+\frac{1}{3!}\frac{1}{(z-1)^3}+ \cdots$$
wherein we observe that the residue at $z=1$ of $z^2e^{\frac{1}{(z-1)}}$ comes from the 3 terms
$$\begin{align}
&(1)\,\,(z-1)^2 \times \frac{1}{3!}\frac{1}{(z-1)^3}=\frac16 (z-1)^{-1}\\
&(2)\,\,2(z-1) \times \frac{1}{2!}\frac{1}{(z-1)^2}=1 (z-1)^{-1}\\
&(3)\,\, 1 \times \frac{1}{z-1}=1 (z-1)^{-1}
\end{align}$$
Thus the residue is $\frac16+1+1=\frac{13}{6}$.  Therefore, 
$$\begin{align}
\oint_{|z=2|} z^3\bar z e^{\frac{1}{(z-1)}}dz&=\oint_{|z=2|} 4z^2 e^{\frac{1}{(z-1)}}dz\\
&=8\pi i \text{Res}_{z=1}\left(z^2 e^{\frac{1}{(z-1)}}\right)\\
&=8\pi i \frac{13}{6}\\
&=\frac{52\pi i}{3}
\end{align}$$
