How to evaluate $\int_{|z|=2} \frac{|z| e^z}{z^2} dz$? The only thing that I know is that the result should be $\displaystyle 4\pi i$. Could you give me a hint/suggestion?
I thought about using the Residue's theorem, where if $\displaystyle f(z) = \frac{|z|e^z}{z^2}$, it has a 2nd order pole in $0$, but after that, I don't know what to do next.
 A: We have 
$$\int_{|z|=2} \frac{|z| e^z}{z^2} dz=2\int_{|z|=2} \frac{e^z}{z^2} dz.\tag{1}$$
Note that, in $\{|z|<2\}$, $z=0$ is the only pole of $\frac{e^z}{z^2}$. By Residue theorem, we have 
$$\int_{|z|=2} \frac{e^z}{z^2} dz=2\pi i Res(\frac{e^z}{z^2}, 0).\tag{2}$$
Note that
$$\frac{e^z}{z^2}=\frac{1}{z^2}\left(1+z+\frac{z^2}{2!}+\cdots\right)
=\frac{1}{z^2}+\frac{1}{z}+\frac{1}{2!}+\cdots.$$
Hence, $Res(\frac{e^z}{z^2}, 0)=1$. 
Combining this with $(1)$ and $(2)$, we obtain 
$$\int_{|z|=2} \frac{|z| e^z}{z^2} dz=2\cdot 2\pi i=4\pi i.$$
A: Hint: this is a line integral in the complex plane. Parametrize $z$ as 
$$
z(t) = 2 e^{i t}
$$
With $t\in [0, 2\pi)$. Then use
$$
\int_\gamma f(z) dz = \int_0^{2\pi} dt f(z(t)) \dot{z}(t)
$$
A: Use DERIVATIVES OF HIGH ORDER FORMULA solve this problem as followed can be quicker:
$$f(\zeta)^{(n)}=\frac{n!}{2\pi i}\oint_{\lvert z \rvert=2}\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta$$
This Formula can be derived by Cauchy Integral Formula
$$\oint_{\lvert z \rvert=2}\frac{\lvert z \rvert e^z}{z^2}dz=2 \oint_{\lvert z \rvert=2}\frac{e^z}{z^2}dz=2*\frac{2\pi i}{1!}*(e^z)^{(1)}\bigg|_{z=0}=4\pi i$$
Note:
$$(e^z)^{(1)}\bigg|_{z=0}=\frac{1!}{2\pi i}\oint_{\lvert z \rvert=2}\frac{e^{z}}{(z-0)^{1+1}}dz$$
