Example 6. Expand the function in powers of the difference $z-3$

$$
f(z)=\frac{1}{3-2 z}
$$
Solution. Let's transform the given function as follows:

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

By substituting $z$ with $\frac{2}{3}(z-3)$ in the expansion (12), we get

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

This series converges under the condition

$$
\left|\frac{2}{3}(z-3)\right|<1
$$

or $|z-3|<\frac{3}{2}$, i.e., the radius of convergence of the series $R=\frac{3}{2}$.