Example 5. Expand the function

$$
f(z)=\frac{z}{z^{2}-2 z-3}
$$

into a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series.
Solution. Let's decompose the given function into partial fractions:

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

Transform the right-hand side as follows:

$$
f(z)=\frac{1}{4} \frac{1}{1+z}-\frac{1}{4} \frac{1}{1-\frac{2}{3}}
$$

Using the expansion (12) of the function $\frac{1}{1+z}$, we get

$$
\begin{aligned}
f(z) & =\frac{1}{4}\left(1-z+z^{2}-z^{3}+\ldots\right)-\frac{1}{4}\left(1+\frac{z}{3}+\frac{z^{2}}{9}+\ldots\right)= \\
& =\frac{1}{4}\left(-\frac{4}{3} z+\frac{8}{9} z^{2}-\frac{28}{27} z^{3}+\ldots\right)=-\frac{z}{3}+\frac{2}{3^{2}} z^{2}-\frac{7}{3^{3}} z^{3}+\ldots
\end{aligned}
$$

The nearest singular point to the point $z_{0}=0$ of the given function is the point $z=-1$. Therefore, the radius of convergence of the obtained series is $R=1$.