Example 12. Expand into a Laurent series in the annulus $0<|z-1|<2$ the function

$$
f(z)=\frac{1}{\left(z^{2}-1\right)^{2}}
$$
Solution: The first method. The function $f(z)=\frac{1}{\left(z^{2}-1\right)^{2}}$ is analytic in the annulus $0<|z-1|<2$, i.e., $n>-3$. Applying formula (2) from § 6 for the derivative of any order of an analytic function, we get

$$
\begin{aligned}
& c_{n}=\frac{1}{2 \pi i} \int_{\Gamma} \frac{\frac{1}{(z+1)^{2}}}{(z-1)^{n+3}} d z=\left.\frac{1}{(n+2)!} \frac{d^{n+2}}{d z^{n+2}}\left[\frac{1}{(z+1)^{2}}\right]\right|_{z=1}= \\
& =\left.\frac{1}{(n+2)!} \cdot \frac{(-1)^{n}(n+3)!}{(z+1)^{n+4}}\right|_{z=1}=\frac{(-1)^{n}(n+3)}{2^{n+4}}
\end{aligned}
$$

For $n=-2,-1,0,1,2, \ldots$ we have

$$
c_{n}=\frac{(-1)^{n}(n+3)}{2^{n+4}}
$$

The Laurent series for the given function in the annulus $0<|z-1|<2$ will be

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

or

$$
\frac{1}{\left(z^{2}-1\right)^{2}}=\frac{1}{4} \frac{1}{(z-1)^{2}}-\frac{1}{4} \frac{1}{z-1}+\frac{3}{16}-\frac{1}{8}(z-1)+\frac{5}{64}(z-1)^{2}-\frac{3}{64}(z-1)^{3}+\ldots
$$

The second method. We need to represent $f(z)$ as a sum of powers (positive and negative) of the difference $(z-1)$: Transform the given function as follows:

$$
\begin{aligned}
f(z) & =\frac{1}{\left(z^{2}-1\right)^{2}}=\frac{1}{4}\left(\frac{1}{z-1}-\frac{1}{z+1}\right)^{2}= \\
& =\frac{1}{4} \frac{1}{(z-1)^{2}}-\frac{1}{4} \frac{1}{z-1}+\frac{1}{4} \frac{1}{z+1}+\frac{1}{4} \frac{1}{(z+1)^{2}}
\end{aligned}
$$

The first two terms on the right side of (24) are already in the required form, as they represent powers of the difference $(z-1)$.

The last two terms can be written as

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

Applying formula (12), and then formula (11) with $\alpha=-2$, we get

$$
\begin{aligned}
\frac{1}{z+1} & =\frac{1}{2}\left[1-\frac{z-1}{2}+\left(\frac{z-1}{2}\right)^{2}-\left(\frac{z-1}{2}\right)^{3}+\cdots\right] \\
\frac{1}{(z+1)^{2}}= & \frac{1}{4}\left[1-2 \cdot \frac{z-1}{2}+\frac{-2(-2-1)}{2!}\left(\frac{z-1}{2}\right)^{2}+\right. \\
& \left.+\frac{-2(-2-1)(-2-2)}{3!}\left(\frac{z-1}{2}\right)^{3}+\cdots\right]
\end{aligned}
$$

Substituting (25) and (26) into (24), we find

$$
\begin{aligned}
\frac{1}{\left(z^{2}-1\right)^{2}} & =\frac{1}{4} \frac{1}{(z-1)^{2}}-\frac{1}{4} \frac{1}{z-1}+ \\
& +\frac{1}{8}\left[1-\frac{z-1}{2}+\left(\frac{z-1}{2}\right)^{2}-\left(\frac{z-1}{2}\right)^{3}+\ldots\right]+ \\
& +\frac{1}{16}\left[1-(z-1)+\frac{3}{2^{2}}(z-1)^{2}-\frac{4}{2^{3}}(z-1)^{3}+\ldots\right]
\end{aligned}
$$

or

$$
\begin{aligned}
\frac{1}{\left(z^{2}-1\right)^{2}}=\frac{1}{4} \frac{1}{(z-1)^{2}}-\frac{1}{4} \frac{1}{z-1} & +\frac{3}{16}-\frac{1}{8}(z-1)+ \\
& +\frac{5}{64}(z-1)^{2}-\frac{3}{64}(z-1)^{3}+\cdots
\end{aligned}
$$