Evaluate the integral of $\frac{e^{−z}}{(z^2)}$ around the circle |z| = 2 Not too sure how to approach this exactly. I think the biggest problem is working out whether or not the function is holomorphic, because from there I can use the appropriate method, but I would appreciate any hints about that too.
 A: The integrand is holomorphic (I just checked), which you can check using the C-R equations.
Method 1: Integration by substitution
$$\int_C \frac{e^{-z}}{z^2}dz,$$ where $z=2e^{i\theta}$ and $\theta\in[0,2\pi]$. Then $dz=2ie^{i\theta}d\theta$, so that
$$\int_0^{2\pi}\frac{e^{-2e^{i\theta}}}{4e^{2i\theta}}2ie^{i\theta}d\theta=\frac{i}{2}\int_0^{2\pi}e^{-2e^{i\theta}-i\theta}d\theta=\frac{i}{2}(-4\pi)=-2\pi i.$$
I used Mathematica to evaluate the last definite integral, but it seems tricky to perform by hand.
Method 2: Residue Theorem
The residue of the integrand at $z=0$ is $-1$, so 
$$\int_C\frac{e^{-z}}{z^2}dz=2\pi i\sum_{\text{residues}}=2\pi i(-1)=-2\pi i,$$
as expected already.
Method 3: Cauchy's Integral Theorem (differentiation formula)
Using the formula $$f^{(n)}(a)=\frac{1}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz,$$
with $f(z)=e^{-z}$, $a=0$ and $n=1$, then $$\int_C\frac{e^{-z}}{z^2}dz=2\pi i f'(0)=-2\pi i.$$
As pointed out in a previous answer, Method 3 is probably the most straightforward approach, depending on taste.
A: Hint: Is the function $f(z)=e^{-z}$ holomorphic?
By Cauchy's theorem for derivatives
$$\int_{|z|=2}\frac{e^{-z}}{z^2}dz=f'(0)2\pi i=-2\pi i$$
