How to find $\frac{f(z)}{z-a}$ I hope that you can help me to find some residues. I know two ways to find the residue in a value $a \in \mathbb{C}$:


*

*Straight forward calculation: $ \int_{C(a,\epsilon)^+} f(z) dz$ 

*Rewriting a function end using the equality $\frac 1{2 \pi i}Res_{z=a}\frac{f(z)}{z-a}\ = \ f(a)$


Now how can I get $Res_{z=0}\frac{e^z}{z^2} $ ? The second trick above doesn't work. So I tried to find the integral:
$$
\int_{C(0,\epsilon)} \frac{e^z}{z} dz \ = \ 
\int_{C(0,\epsilon)} \frac{e^{\epsilon e^{it}}}{\epsilon e^{it}} \cdot i\epsilon e^{it}dz 
\ = \ i \int_{C(0,\epsilon)} e ^ {re^{it}-it}dt
$$
I don't know how to solve this, so I hope that you can give me a trick to do so.
 A: Use series expansion of $e^z$ to get Laurent expansion of your function, you have:
$$e^{z}=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
So:
$$\frac{e^{z}}{z^2}=\sum_{n=0}^{\infty}\frac{z^{n-2}}{n!}=\frac{1}{z^2}+\frac{1}{z}+\sum_{n=2}^{\infty}\frac{z^{n-2}}{n!}$$
$\displaystyle \sum_{n=2}^{\infty}\frac{z^{n-2}}{n!}$ is analytic (entire function), so $Res_{z=0}\frac{e^z}{z^2}=1$.
A: I guess you might call the following a "trick":
Lemma Let $D\subseteq \mathbb{C}$ be open and let $z_{0}\in D$.  Assume that $f$ is holomorphic in $D\setminus\{z_{0}\}$ and has a pole of order $m$ in $z_{0}$. Then
$$
Res_{z_{0}}(f)=\frac{1}{(m-1)!}\lim_{z\to z_{0}}((z-z_{0})^m f(z))^{(m-1)},
$$
where $((z-z_{0})^m f(z))^{(m-1)}$ means the $(m-1)-$th derivative of $(z-z_{0})^m f(z)$.
The proof is almost immediate: write down the Laurent development at $z_{0}$ of $f$ and look at what is $((z-z_{0})^m f(z))^{(m-1)}$ in terms of this development.
Now, apply the result to your case, with $f(z):=e^z/z^2$, which has a pole of order $2$ at $z_{0}=0$.
You might also find the following result useful:
Lemma Let $D\subseteq \mathbb{C}$ be open and let $z_{0}\in D$.  Assume that $g$ is holomorphic in $z_{0}$, while $f$ is holomorphic in $D\setminus\{z_{0}\}$ and has a simple pole in $z_{0}$. Then
$$
Res_{z_{0}}(fg)=g(z_{0})Res_{z_{0}}(f)
$$
Again, the proof of this fact is clear: write the Laurent development of $f$ and $g$ in $z_{0}$ and look at what is the term indexed by $-1$ in the Laurent development of $fg$.
A: If you have a function holomorphic on some annulus we have a Laurent expansion
$$f(z) = \sum_{n= \infty}^{\infty} a_n (z-z_0)^n$$
where
$$a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz$$
where $\gamma$ is some closed curve in your annulus.
We have the the residue at $z_0$ is $Res_{z=z_0} f(z) = a_{-1}$. That is the residue is just the coefficient of $z^{-1}$ in our Laurent series. We can thus compute the residue with the integral formula or find the Laurent series and look at the coffient $a_{-1}$.
So remember all the residue is a some coefficient in the Laurent series. Sometimes it is hard to compute the Laurent series center at a given point and the integral computation is better to do. Other times it is easier to just find the Laurent series, which it is in your cases as the other answers have pointed out.
