# Coefficient of Multiplication of Two Laurent Series

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*}
• Yes, I do understand that I wish to know whether there is a general formula or mechanism to evaluate the series at the end in terms of $\sum_{-\infty}^{\infty} c_m z^m$. Commented Nov 22, 2019 at 8:58
• I did find the general expression. If it is $\sum_{-\infty}^{\infty}$, then $c_n=-e$ if n$\geq 0$ and $c_n= -e+\sum_{i=0}^{j-1} \frac{1}{i!}$ if $-n=2j-1 or 2j$. But I want to know if there is some other mechanism to directly obtain it Commented Nov 22, 2019 at 8:58
• I don't think there is any alternative way to obtain such a closed form expression other than brute force manipulation, due to the coefficients taking the form $\sum_{s=-\infty}^{\infty}a_{n-s}b_{s}$. Commented Nov 23, 2019 at 8:49