Expand a function into a Laurent series about a point? Take the function $f(z)=(z^2+3z+2)e^\frac{1}{z+1}$
We want to expand this into its Laurent series about $z_0$=-1.
Alright, so I'm a little confused. This converges everywhere but -1, which throws me off. My first thought is to just start expanding it using properties that I know. So, I think we have the following:
$e^z=1+z+\frac{1}{2!}z^2+\frac{1}{3!}z^3+...$
$\frac{1}{1-(-z)}=1-z+z^2-z^3+z^4-...$
But I'm not sure how to expand the term $e^\frac{1}{z+1}$. It seems like a composition of expansions. Also, can I use a substitution to account for the point -1? This seems like it's going to turn out very messily. I'd appreciate any guidance.
 A: Hint. You may just write, for $z\neq-1$,
$$e^{\large\frac1{z+1}}=1+\frac1{(z+1)}+\frac1{2!(z+1)^2}+\cdots+\frac1{n!(z+1)^n}+\cdots
$$ and 
$$
(z^2+3z+2)=(z+1)^2+(z+1)
$$ then expand
$$
\left((z+1)^2+(z+1)\right)\times\left(1+\frac1{(z+1)}+\frac1{2!(z+1)^2}+\cdots+\frac1{n!(z+1)^n}+\cdots\right)
$$ Can you take it from here?

Set $X=z+1$, you get
$$
\begin{align}
(X^2+X)\sum_{0}^{\infty}\frac1{n!}\frac1{X^n}&=\sum_{0}^{\infty}\frac1{n!}\frac1{X^{n-2}}+\sum_{0}^{\infty}\frac1{n!}\frac1{X^{n-1}}\\\\
&=\sum_{-2}^{\infty}\frac1{(n+2)!}\frac1{X^{n}}+\sum_{-1}^{\infty}\frac1{(n+1)!}\frac1{X^{n}}\\\\
&=\frac1{X^{-2}}+\sum_{-1}^{\infty}\left(\frac1{(n+2)!}+\frac1{(n+1)!}\right)\frac1{X^{n}}\\\\
&=\frac1{X^{-2}}+\sum_{-1}^{\infty}\frac1{(n+1)!}\left(\frac1{(n+2)}+1\right)\frac1{X^{n}}\\\\
&=\frac1{X^{-2}}+\sum_{-1}^{\infty}\frac{n+3}{(n+2)!}\frac1{X^{n}}\\\\
&=\sum_{-2}^{\infty}\frac{n+3}{(n+2)!}\frac1{X^{n}}
\end{align}
$$ Then the Laurent series expansion of your function is 

$$
(z^2+3z+2)e^{\large \frac{1}{z+1}}=\sum_{-2}^{\infty}\frac{n+3}{(n+2)!}\frac1{(z+1)^{n}},\quad z\neq-1.
$$

