# Laurent series expansion of $f(z)=\frac{1}{(z-1)^2(z+1)^2}$

Let $$f(z)=\frac{1}{(z-1)^2(z+1)^2}$$. While trying to expand this function into the Laurent series, convergent in $$P(0,1,2):=\lbrace z\in\mathbb{C}:1<|z|<2\rbrace$$, a few questions popped into my mind.

1. We can write $$f(z)=\frac{1}{4}\left(\frac{1}{(z-1)^2}+\frac{1}{(z+1)^2}\right)$$. Both functions inside parentheses are complex derivatives of functions which have immediate Laurent series expansion: $$\frac{1}{1-z}$$ and $$-\frac{1}{z+1}$$. Now, can we differentiate the obtained series term by term to get the desired expansion of $$f$$? If so, is it because the Laurent series is convergent almost uniformly?
2. Could someone verify that the Laurent series of $$f$$ is convergent in $$P(0,1,\infty)$$?
• Here is a related problem. Commented Feb 5, 2013 at 10:25
• I just realized I wrote something which is not true ($f$ is not the sum of the two functions above). But maybe I won't edit the post, since this mistake led me to a question I wanted to ask anyway. Commented Feb 5, 2013 at 11:56

Hint:

$$z\in P(0,1,2)\Longleftrightarrow \frac{1}{2}<\frac{1}{|z|}<1\Longleftrightarrow z\in P\left(0,\frac{1}{2},1\right)$$

so using the well-known developments

$$\frac{1}{1-z}=1+z+z^2+z^3+\ldots\;\;,\;\;\frac{1}{1+z}=1-z+z^2-z^3+\ldots$$

we get

$$\frac{1}{(z-1)^2(z+1)^2}=\frac{1}{z^4}\frac{1}{\left(1-\frac{1}{z}\right)^2}\frac{1}{\left(1+\frac{1}{z}\right)^2}=$$

$$=\frac{1}{z^4}\left(1+\frac{1}{z}+\frac{1}{z^2}+\ldots\right)^2\left(1-\frac{1}{z}+\frac{1}{z^2}-\ldots\right)^2=\ldots$$

Can you take it from here?

$$\frac{1}{(z-1)^2(z+1)^2}=\frac{1}{(z^2-1)^2}=\frac{1}{z^4}\frac{1}{\left(1-\frac{1}{z^2}\right)^2}=\frac{1}{z^4}\left(1+\frac{1}{z^2}+\frac{1}{z^4}+\ldots\right)^2=$$

$$=\frac{1}{z^4}\left(1+\frac{2}{z^2}+\frac{3}{z^4}+\frac{4}{z^6}+\ldots+\frac{n}{z^{2n-2}}+\ldots\right)=\sum_{n=1}^\infty\frac{n}{z^{2n+2}}$$

2) Yes, since the only poles of $f$ are at $z = \pm 1$. (I.e., $f$ is holomorphic on every annulus $1 < |z| < r$ for $r > 1$, and on that annulus you will get a convergent Laurent series.)