Example 9. Find the region of convergence of the series

$$
\sum_{n=1}^{\infty} \frac{\sin i n}{(z+i)^{n}}
$$
Solution. We have

$$
c_{-n}=\sin i n=i \operatorname{sh} n, \quad c_{-n-1}=i \operatorname{sh}(n+1).
$$

Therefore,

$$
r=\lim _{n \rightarrow \infty} \frac{|i \operatorname{sh}(n+1)|}{|i \operatorname{sh} n|}=\lim _{n \rightarrow \infty} \frac{\operatorname{sh}(n+1)}{\operatorname{sh} n}=\lim _{n \rightarrow \infty} \frac{e^{n+1}-e^{-n-1}}{e^{n}-e^{-n}}=\lim _{n \rightarrow \infty} \frac{e-e^{-2 n-1}}{1-e^{-2 n}}=e
$$

Thus, the series converges in the region $|z+i|>e$, i.e., outside the circle with center at the point $z_{0}=-i$ and radius $e$.

## Problems for Independent Solution

Determine the region of convergence of the following series:
229. $\sum_{n=1}^{\infty} \frac{1}{(1-i)^{n} z^{n}}$. 230. $\sum_{n=1}^{\infty} \frac{(\sqrt{2}+i \sqrt{2})^{n}}{z^{n}}$.
231. $\sum_{n=1}^{\infty} \frac{z^{-n}}{\cos i n}$.
232. $\sum_{n=1}^{\infty} e^{n}(i z)^{-n}$.
233. $\sum_{n=1}^{\infty} \frac{1}{4^{n}(z+1)^{n}}$.
234. $\sum_{n=1}^{\infty} \frac{n 2^{-n}}{(z-2-i)^{n}}$.
235. $\sum_{n=1}^{\infty} \frac{3^{n}+1}{(z+2 i)^{n}}$.
236. $\sum_{n=1}^{\infty} \frac{(z+1-i)^{-n}}{n+i}$.

## Series of the Form

$$
\begin{gathered}
\sum_{n=-\infty}^{\infty} c_{n}\left(z-z_{0}\right)^{n}=\sum_{n=1}^{\infty} \frac{c_{-n}}{\left(z-z_{0}\right)^{n}}+\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n}=\ldots+\frac{c_{-n}}{\left(z-z_{0}\right)^{n}}+\ldots \\
\ldots+\frac{c_{-1}}{z-z_{0}}+c_{0}+c_{1}\left(z-z_{0}\right)+\ldots+c_{n}\left(z-z_{0}\right)^{n}+\ldots
\end{gathered}
$$

converges in the region where the series

$$
\begin{aligned}
\sum_{n=1}^{\infty} \frac{c_{-n}}{\left(z-z_{0}\right)^{n}} & =\frac{c_{-1}}{z-z_{0}}+\frac{c_{-2}}{\left(z-z_{0}\right)^{2}}+\ldots \\
\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n} & =c_{0}+c_{1}\left(z-z_{0}\right)+c_{2}\left(z-z_{0}\right)^{2}+\ldots
\end{aligned}
$$

converge. Let the series (20) converge in the region $\left|z-z_{0}\right|>r$, i.e., outside the circle with center at the point $z=z_{0}$ and radius $r$, and the series (21) in the circle $\left|z-z_{0}\right|<R$. Then:
1) If $r \geq R$, the series (19) diverges everywhere;
2) If $r<R$, the series (19) converges in the annulus $r<\left|z-z_{0}\right|<R$. Here $r \geqslant 0,0<R<+\infty$.