Example 14. Consider various Laurent series expansions of the function

$$
f(z)=\frac{2 z+1}{z^{2}+z-2}
$$

taking $z_{0}=0$.
Solution. The function $f(z)$ has two singular points: $z_{1}=-2$ and $z_{2}=1$. Therefore, there are three "rings" centered at the point $z_{0}=0$, in each of which $f(z)$ is analytic:
a) $\mathrm{kpyr}|z|1$. Therefore, we transform $f(z)$ as follows:

$$
f(z)=\frac{1}{2} \frac{1}{1+\frac{z}{2}}+\frac{1}{z} \frac{1}{1-\frac{1}{z}}
$$

Applying formula (12), we get

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

This series converges for $\left|\frac{1}{z}\right|1$.

Substituting (30) and (32) into (31), we find

$$
\begin{aligned}
\frac{2 z+1}{z^{2}+z-2} & =\frac{1}{2}-\frac{z}{4}+\frac{z^{2}}{8}-\frac{z^{3}}{16}+\ldots+\frac{1}{z}+\frac{1}{z^{2}}+\ldots= \\
& =\ldots+\frac{1}{z^{2}}+\frac{1}{z}+\frac{1}{2}-\frac{z}{4}+\frac{z^{2}}{8}-\frac{z^{3}}{16}+\ldots
\end{aligned}
$$

or

$$
\frac{2 z+1}{z^{2}+z-2}=\sum_{n=1}^{\infty} \frac{1}{z^{n}}+\frac{1}{2} \sum_{n=0}^{\infty} \frac{z^{n}}{2^{n}}
$$

b) Expansion for $|z|>2$. The series (30) for the function $\frac{1}{1+z / 2}$ diverges for $|z|>2$, while the series (32) for the function $\frac{1}{1-1 / z}$ will converge, since if $|z|>2$, then certainly $|z|>1$.

We represent the function $f(z)$ in the following form:

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

Using formula (12), we get

$f(z)=\frac{1}{z}\left(1-\frac{2}{z}+\frac{4}{z^{2}}-\frac{8}{z^{3}}+\cdots+1+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\ldots\right)=\frac{1}{z}\left(2-\frac{1}{z}+\frac{5}{z^{2}}-\frac{7}{z^{3}}+\cdots\right)$, or

$$
\frac{2 z+1}{z^{2}+z-2}=\frac{2}{z}-\frac{1}{z^{2}}+\frac{5}{z^{3}}-\frac{7}{z^{4}}+\cdots
$$

This example shows that for the same function $f(z)$, the Laurent series, in general, has a different form for different rings.