problems with singularity $0$ of $\int_{W} \frac{e^{\frac{1}{z}}}{(z-3)^3} dz$. I have the complex integral 
\begin{equation*}
\int_{W} \frac{e^{\frac{1}{z}}}{(z-3)^3} dz 
\end{equation*}
where $W$ is a circle with radius $6$ and centered at $0$.
Obviously we have two singularities, one in $0$ and one in $3$. I only get confused by the singularity in $0$. I don't clearly see what $\lim_{z\rightarrow0}$ of $zf(z)$ is. the limit $ze^\frac{1}{z}$ bothers me a lot and l'hopital does not provide a clear result.
Any tips?
Kees
 A: You have an essential singularity at $0$, so $z^k\cdot f(z)$ has no limit at $0$, for any $k$.
In principle, you could develop $\frac{1}{(z-3)^3}$ into a geometric series around $0$, multiply that with the series for $e^{1/z}$, and find a series representation of the residue at $0$ in that way. That would however be a lot of work, where mistakes can easily be made.
But you can also use Cauchy's integral theorem to shift the contour. The integral
$$\int_{\lvert z\rvert = R} \frac{e^{1/z}}{(z-3)^3}\,dz$$
is for $R > 3$ independent of $R$. For $R \geqslant 6$ and $\lvert z\rvert = R$, we have $\lvert z-3\rvert \geqslant \lvert z\rvert - 3 \geqslant \frac{R}{2}$, and since $\lvert e^{1/z}\rvert \leqslant e^{1/\lvert z\rvert}$, we find
$$\Biggl\lvert \int_{\lvert z\rvert = R} \frac{e^{1/z}}{(z-3)^3}\,dz\Biggr\rvert \leqslant 2\pi R\frac{8e^{1/R}}{R^3} \leqslant \frac{16\pi e^{1/6}}{R^2}$$
by the standard estimate, so
$$\lim_{R\to\infty} \int_{\lvert z\rvert = R} \frac{e^{1/z}}{(z-3)^3}\,dz = 0.$$
Since the integral doesn't depend on $R$,
$$\int_{\lvert z\rvert = R} \frac{e^{1/z}}{(z-3)^3}\,dz = 0$$
for all $R > 3$.
More conceptually, if you are familiar with the Riemann sphere $\widehat{\mathbb{C}}$, we can view the circle $\lvert z\rvert = 6$ as the boundary curve of $G = \widehat{\mathbb{C}} \setminus \{ z : \lvert z\rvert \leqslant 6\}$, and the residue theorem then says
$$\int_{\lvert z\rvert = R} \frac{e^{1/z}}{(z-3)^3}\,dz = -2\pi i \sum_{\zeta\in G} \operatorname{Res}\biggl(\frac{e^{1/z}}{(z-3)^3};\zeta\biggr) = -2\pi i \operatorname{Res}\biggl(\frac{e^{1/z}}{(z-3)^3}; \infty\biggr).$$
The minus sign comes from the orientation of the circle, $G$ lies to the right of the circle. Since by definition
$$\operatorname{Res}(f(z);\infty) = \operatorname{Res}\biggl(-\frac{1}{z^2}f\biggl(\frac{1}{z}\biggr); 0\biggr),$$
we compute
\begin{align}
\operatorname{Res}\biggl(\frac{e^{1/z}}{(z-3)^3};\infty\biggr)
&= \operatorname{Res}\Biggl(-\frac{1}{z^2}\frac{e^z}{\bigl(\frac{1}{z}-3\bigr)^3}; 0\Biggr)\\
&= \operatorname{Res}\biggl(-\frac{ze^z}{(1-3z)^3};0\biggr)\\
&= 0.
\end{align}
A: $$\int_{W} \frac{e^{\frac{1}{z}}}{(z-3)^3} dz $$
$$=2\pi i[Res(f,0)+Res(f,3)]=-2\pi iRes(f,\infty)=0$$where , $f(z)=\frac{e^{\frac{1}{z}}}{(z-3)^3} $. Since, sum of residues at finite poles and the residue at $\infty$ is equal to zero.
Since,  $Res(f,\infty)=-\lim_{z\to \infty}zf(z)$ , as $f(\infty)=0$
$=-\lim_{z\to \infty}\frac{z}{(z-3)^3}e^{1/z}=0$.
