# Multiplying Laurent series

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 slight 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 it's 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: 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.

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Absolute convergence is your friend. –  copper.hat Apr 12 '13 at 22:47
Thank you copper.hat. I'm thinking about how to use that. –  bryanj Apr 12 '13 at 23:29
Here is a related problem. –  Mhenni Benghorbal Apr 12 '13 at 23:37
I want to write an answer, but am stuck for time... –  copper.hat Apr 13 '13 at 0:00
@ Mhenni Benghorbal - Thanks for the link! –  bryanj Apr 13 '13 at 1:19

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 $c_k = \sum_l f_l g_{k-l}$.

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That was awesome! Thanks copper.hat. I shouldn't really ask for more, but...did you notice an obvious flaw in my reasoning? –  bryanj Apr 13 '13 at 16:22
@bryanj: Your analysis looks good and is basically the same idea, but simpler. I wanted to show that it can be done directly with the series themselves (albeit with more grunt work). You know that $c(z)= f(z)g(z)$ is analytic on the intersection of their respective annuli and so has a (unique!) Laurent expansion on this intersection. Your computation computes the $c_k$. The annulus comes into the computation implicitly with the choice of contour $C$ which must lie inside $R_- < r < R_+$ as above. –  copper.hat Apr 13 '13 at 16:59
Thanks again - this whole thing was very helpful. –  bryanj Apr 13 '13 at 17:29