I was solving the following problem from Bak & Newmans Complex Analysis (Chp. 9 # 12b):
Find the Laurent series (in powers of z) for
$$\frac{1}{z(z-1)(z-2)}$$
in the open annulus $1 < |z| < 2$.
I worked out the solution in a couple of ways. The first way used partial fraction decomposition then plugged in the Laurent series for $\frac{1}{z-1}$ and $\frac{1}{z-2}$.
After the first method seemed to work out, I thought I'd try to get the answer without using the partial fraction decomposition, but instead directly multiply (like a Cauchy product) the Laurent series for $\frac{1}{z-1}$ and $\frac{1}{z-2}$.
The second method gave the same answer as the first method, but I feel like I did some illegal sleight of hand. Taking the Cauchy product of power series involves evaluating a finite sum to get a coefficient. But performing an analogous operation for Laurent series can involve trying to sum an infinite series to find a coefficient.
Are there any useful conditions that guarantee that this kind of formal product does what you expect it to – meaning that the sum/series that defines each coefficient converges and that the resulting Laurent series itself converges to the product of the values of the original two series?
I was guessing that maybe you might be allowed to multiply two Laurent series on an annulus (if there is one) where they both converge. I thought this might work for the following reason:
Let $f(z) = \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$, $g(z) = \sum_{n=-\infty}^{\infty}b_n(z-z_0)^n$, and say they both converge on the same annulus around some point $z_0 \in \mathbb{C}$. The product $f(z)g(z)$ is also analytic in the annulus, and so should have a Laurent series in that domain where the coefficient $c_k$ of $z^k$ is given by
$$c_k = \frac{1}{2\pi i}\int_C{\frac{f(z)g(z)}{(z-z_0)^{k+1}}}dz $$
Then you replace $g(z)$ inside the integral with its Laurent series in powers of $(z-z_0)$:
$$c_k = \frac{1}{2\pi i}\int_C{\frac{f(z)}{(z-z_0)^{k+1}} \sum_{n=-\infty}^{\infty}b_n(z-z_0)^n dz } $$
Does uniform convergence allow us to interchange the integral and the sum? If so then it seems like each term is then a product $b_n a_{k-n}$.
EDIT: Here are more details, which I'm hoping are correct.
$$c_k = \frac{1}{2\pi i}\int_C{ \frac{f(z)}{(z-z_0)^{k+1}} \left( \sum_{n=-1}^{-\infty}{b_n(z-z_0)^n } + \sum_{n=0}^{\infty}{b_n(z-z_0)^n } \right) dz} $$
$$c_k = \frac{1}{2\pi i}\int_C{ \frac{f(z)}{(z-z_0)^{k+1}} \sum_{n=-1}^{-\infty}{b_n(z-z_0)^n } dz} + \frac{1}{2\pi i}\int_C{ \frac{f(z)}{(z-z_0)^{k+1}} \sum_{n=0}^{\infty}{b_n(z-z_0)^n } dz} $$
Both integrands converge uniformly on the circle $C$ since the series do and $\frac{f(z)}{(z-z_0)^{k+1}}$ is analytic on $C$ (and so bounded). So we can interchange the integral and the limit of the sequence of partial sums to get
$$c_k = \sum_{n=-1}^{-\infty}{b_n \frac{1}{2\pi i}\int_C{ \frac{f(z)}{(z-z_0)^{k – n + 1}} } dz}
+ \sum_{n=0}^{\infty}{b_n \frac{1}{2\pi i}\int_C{ \frac{f(z)}{(z-z_0)^{k – n +1 }} } dz} $$
$$c_k = \sum_{n=-1}^{-\infty}{b_n a_{k-n}}
+ \sum_{n=0}^{\infty}{b_n a_{k-n}} $$
$$c_k = \sum_{n=-\infty}^{-\infty}{b_n a_{k-n}}$$
I feel like I must have made a mistake somewhere in the last 5 or so lines, because the (to me strange) conclusion I'm reaching is that if two Laurent series both converge on the same annulus then the sum $\sum_{n=-\infty}^{-\infty}{b_n a_{k-n}}$ converges for each $k$. But the sum has nothing to do with the annulus…
Thanks again folks.
Best Answer
You can obtain the desired result using absolute convergence and some observations about the annulus of convergence of Laurent series.
If $\{x_n\}_{n \in \mathbb{Z}}$, then let $R_+(\{x_n\}) = ( \limsup_{n \to \infty} \sqrt[n]{|x_n|} )^{-1}$, and $R_-(\{x_n\}) = \limsup_{n \to \infty} \sqrt[n]{|x_{-n}|}$.
Two relevant results (all summations are over $\mathbb{Z}$):
(i) If $\sum_m \sum_n |a_{m,n}| < \infty$, then the summation can be rearranged. In particular, $\sum_m \sum_n a_{m,n} = \sum_k \sum_l a_{l,k-l}$. (Since $\phi(m,n) = (m,m+n)$ is a bijection of $\mathbb{Z}^2$ to $\mathbb{Z}^2$.)
(ii) If $\sum_{n} |x_n| < \infty$, then $R_+(\{x_n\}) \ge 1$. Similarly, $R_-(\{x_n\}) \le 1$.
Let $f(z) = \sum_n f_n (z-z_0)^n$, $g(z) = \sum_n g_n (z-z_0)^n$. Let $R_+ = \min(R_+(\{f_n\}),R_+(\{g_n\}))$, $R_- = \max(R_-(\{f_n\}),R_-(\{g_n\}))$.
Choose $R_-<r < R_+$. Then $\sum_m |f_m| r^m$ and $\sum_n |g_n| r^n$ are absolutely convergent sequences, and so $\sum_m \sum_n |f_m||g_n| r^{m+n} < \infty$. From (i) we have $\sum_m \sum_n |f_m||g_n| r^{m+n} = \sum_k (\sum_l |f_l| |g_{k-l}|) r^k < \infty$.
If we let $c_k = \sum_l f_l g_{k-l}$, this gives $\sum_k |c_k| r^k < \infty$, and so $R_+(\{c_kr^k\}) = \frac{1}{r} R_+(\{c_k \}) \ge 1$, or, in other words, $R_+(\{c_k \}) \ge r$. Since $r<R_+$ was arbitrary, it follows that $R_+(\{c_k \}) \ge R_+$. The same line of argument gives $R_-(\{c_k \}) \le R_-$.
Hence we have $c(z) = f(z)g(z) = \sum_n c_n (z-z_0)^k$ on $R_- < |z-z_0| < R_+$, where $c_k = \sum_l f_l g_{k-l}$.