Example 11. Determine the region of convergence of the series

$$
\sum_{n=1}^{\infty} \frac{(3+4 i)^{n}}{(z+2 i)^{n}}+\sum_{n=0}^{\infty}\left(\frac{z+2 i}{6}\right)^{n}
$$
Solution. For the series $\sum_{n=1}^{\infty} \frac{(3+4 i)^{n}}{(z+2 i)^{n}}$ we have

$$
c_{-n}=(3+4 i)^{n}, \quad c_{-n-1}=(3+4 i)^{n+1}
$$

Therefore,

$$
r=\lim _{n \rightarrow \infty} \frac{\left|(3+4 i)^{n+1}\right|}{\left|(3+4 i)^{n}\right|}=\lim _{n \rightarrow \infty}|3+4 i|=5
$$

The first series converges in the region $|z+2 i|>5$. For the series $\sum_{n=0}^{\infty}\left(\frac{z+2 i}{6}\right)^{n}$ we have

$$
c_{n}=6^{-n} \quad c_{n+1}=6^{-n-1}
$$

Therefore, the radius of convergence of this power series is

$$
R=\lim _{R \rightarrow \infty} \frac{\left|6^{-n}\right|}{\left|6^{-n-1}\right|}=6
$$

It converges in the region $|z+2 i|<6$.

Thus, $r=5<R=6$. Therefore, the given series converges in the annulus $5<|z+2 i|<6$.

## Problems for Independent Solution

Determine the regions of convergence of the following series:

237. $\sum_{n=1}^{\infty}\left(\frac{1}{n}+i n\right)(z+1+i)^{n}$
238. $\sum_{n=1}^{\infty} \frac{(z-i)^{n}}{2^{n}}$.
239. $\sum_{n=1}^{\infty} \frac{1}{n^{n}(z-2+i)^{n}}+\sum_{n=0}^{\infty}(1+i n)(z-2+i)^{n}$.
240. $\sum_{n=1}^{\infty}\left(\frac{2}{z}\right)^{n}+\sum_{n=0}^{\infty}\left(\frac{z}{4}\right)^{n}$.
241. $\sum_{n=1}^{\infty} \frac{n}{(z+1-i)^{n}}+\sum_{n=0}^{\infty} n(z+1-i)^{n}$.
242. $\sum_{n=1}^{\infty} \frac{1}{z^{n}}+\sum_{n=0}^{\infty} \frac{z^{n}}{2^{n+1}}$.
243. $\sum_{n=1}^{\infty} \frac{\sin i n}{(z-i)^{n}}+\sum_{n=0}^{\infty} \frac{(z-i)^{n}}{n!} \quad(0!=1)$.
244. $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{4} z^{n}}+\sum_{n=1}^{\infty} \frac{z^{n}}{n 2^{n}}$.
245. $\sum_{n=1}^{\infty} \frac{2^{n}-1}{(z+1)^{n}}+\sum_{n=0}^{\infty} \frac{(z+1)^{n}}{(i+n)^{n}}$ :
246. $-\frac{i}{2(z-i)}+\frac{1}{4} \sum_{n=0}^{\infty}(-1)^{n} \frac{(z-i)^{n}}{(2 i)^{n}}$.
247. $\frac{1}{z}+\sum_{n=0}^{\infty} z^{n}$
248. $-\frac{1}{z-1}+\sum_{n=0}^{\infty}(-1)^{n}(z-1)^{n}$.
249. $\sum_{n=1}^{\infty} \frac{a^{n}}{z^{n}}+\sum_{n=0}^{\infty} \frac{z^{n}}{b^{n}}(b \neq 0)$.

The function $f(z)$, which is single-valued and analytic in the annulus $r<\left|z-z_{0}\right|<R$ (the cases where $r=0$ and $R=+\infty$ are not excluded), can be expanded in this annulus into a Laurent series

$$
f(z)=\sum_{n=-\infty}^{\infty} c_{n}\left(z-z_{0}\right)^{n}=\sum_{n=-\infty}^{-1} c_{n}\left(z-z_{0}\right)^{n}+\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n}
$$

where the coefficients $c_{n}$ are found using the formulas

$$
c_{n}=\frac{1}{2 \pi i} \int_{\Gamma} \frac{f(z) d z}{\left(z-z_{0}\right)^{n+1}} \quad(n=0, \pm 1, \pm 2, \ldots)
$$

Here $\Gamma$ is any circle centered at the point $z_{0}$, lying inside the given annulus.

In the formula (22), the series

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

is called the principal part of the Laurent series, and the series

$$
\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n}
$$

is called the regular part of the Laurent series.

In practice, when finding the coefficients $c_n$, one tries to avoid using formulas (23) as they lead to cumbersome calculations. Usually, if possible, ready-made Taylor series expansions of elementary functions are used.