If I have two Laurent Series $\sum_{n=- \infty}^{\infty} a_nz^n$ and $\sum_{n=- \infty}^{\infty} b_nz^n$ what will be the coeefficients $c_n$ if $\sum_{n=- \infty}^{\infty} a_nz^n*\sum_{n=- \infty}^{\infty} b_nz^n$ = $\sum_{n=- \infty}^{\infty} c_nz^n$.
I am solving Complex Analysis by Bak Newman and in Chapter 9 Exercise 9b, I need to find Laurent expansion of $\frac{exp(\frac{1}{z^2})}{z-1}$ about $z=0$, and it is equal to $\sum_{k=0}^{\infty} \frac{1}{k! z^{2k}} * \sum_{l=0}^{\infty} z^n$
To answer your first question, if we write \begin{align*} \sum_{k=-\infty}^{\infty}a_{k}z^{k}\sum_{l=-\infty}^{\infty}b_{l}z^{l}=\sum_{m=-\infty}^{\infty}c_{m}z^{m}, \end{align*} the coefficient of $z^{n}$ on the RHS is given by \begin{align*} &\dots+a_{n+2}b_{-2}+a_{n+1}b_{-1}+a_{n}b_{0}+a_{n-1}b_{1}+a_{n-2}b_{2}+\dots+a_{1}b_{n-1}+a_{0}b_{n}+a_{-1}b_{n+1}+\dots \\ &=\sum_{s=-\infty}^{\infty}a_{n-s}b_{s}. \end{align*} To answer your second question, we first recall the Taylor Series expansions of $e^{z}$ and $1/(1-z)$ for $z\in\mathbb{C}$ about $z=0$. They are \begin{align*} e^{z}&=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\dots=\sum_{k=0}^{\infty}\frac{z^{k}}{k!} \\ \frac{1}{1-z}&=1+z+z^{2}+z^{3}+\dots=\sum_{l=0}^{\infty}z^{l} \end{align*} The Laurent Series of $e^{1/z^{2}}/(1-z)$ is given by \begin{align*} \frac{e^{1/z^{2}}}{1-z}=e^{1/z^{2}}\left(\frac{1}{1-z}\right)=\sum_{k=0}^{\infty}\frac{(1/z^{2})^{k}}{k!}\sum_{l=0}^{\infty}z^{l}=\sum_{k=0}^{\infty}\frac{1}{k!z^{2k}}\sum_{l=0}^{\infty}z^{l}. \end{align*}