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.
- 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?
- Could someone verify that the Laurent series of $f$ is convergent in $P(0,1,\infty)$?
Best Answer
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?
Added: A different approach:
$$\frac{1}{(z-1)^2(z+1)^2}=\frac{1}{(z^2-1)^2}=\frac{1}{z^4}\frac{1}{\left(1-\frac{1}{z^2}\right)^2}=\frac{1}{z^4}\left(1+\frac{1}{z^2}+\frac{1}{z^4}+\ldots\right)^2=$$
$$=\frac{1}{z^4}\left(1+\frac{2}{z^2}+\frac{3}{z^4}+\frac{4}{z^6}+\ldots+\frac{n}{z^{2n-2}}+\ldots\right)=\sum_{n=1}^\infty\frac{n}{z^{2n+2}}$$
Check this thoroughly, please.