Example 4. Find the radius of convergence of the power series

$$
\sum_{n=0}^{\infty}(1+i)^{n} z^{n}
$$
Solution. We find the modulus of the coefficient $c_{n}=(1+i)^{n}$:

$$
\left|c_{n}\right|=\left|(1+i)^{n}\right|=|1+i|^{n}=(\sqrt{2})^{n}=2^{n / 2}
$$

Applying formula (7), we find the radius of convergence of the given power series:

$$
R=\lim _{n \rightarrow \infty} \frac{1}{\sqrt[n]{2^{n / 2}}}=\frac{1}{\sqrt{2}}
$$

## Problems for Independent Solution

Find the radii of convergence of the following power series:

197. $\sum_{n=1}^{\infty} e^{i n} z^{n}$
198. $\sum_{n=1}^{\infty} e^{i \pi / n} z^{n}$
199. $\sum_{n=0}^{\infty}\left(\frac{z}{1-i}\right)^{n}$
200. $\sum_{n=1}^{\infty}\left(\frac{z}{i n}\right)^{n}$.
201. $\sum_{n=1}^{\infty} \operatorname{ch} \frac{i}{n} z^{n}$.
202. $\sum_{n=1}^{\infty}\left(\frac{z}{\ln i n}\right)^{n}$.
203. $\sum_{n=0}^{\infty} i^{n} z^{n}$
204. $\sum_{n=1}^{\infty} \sin \frac{\pi i}{n} z^{n}$
205. $\sum_{n=1}^{\infty} \cos ^{n} \frac{\pi i}{\sqrt{n}} z^{n}$
206. $\sum_{n=1}^{\infty} \frac{z^{n}}{\sin ^{n}(1+i n)} \cdot$ 207. $\sum_{n=0}^{\infty}(n+i) z^{n} . \quad$ 208. $\sum_{n=0}^{\infty} \cos i n \cdot z^{n}$.
207. The series $\sum_{n=0}^{\infty} c_{n} z^{n}$ has a radius of convergence $r$, and the series $\sum_{n=0}^{\infty} c_{n}^{\prime} z^{n}$ has a radius of convergence $\boldsymbol{r}^{\prime}$.

Estimate the radius of convergence $\boldsymbol{R}$ of the following series:
a) $\sum_{n=0}^{\infty}\left(c_{n}+c_{n}^{\prime}\right) z^{n}$
b) $\sum_{n=0}^{\infty}\left(c_{n}-c_{n}^{\prime}\right) z^{n}$;
c) $\sum_{n=0}^{\infty} c_{n} c_{n}^{\prime} z^{n}$
d) $\sum_{n=0}^{\infty} \frac{c_{n}}{c_{n}^{\prime}} z^{n}\left(c_{n}^{\prime} \neq 0\right)$

## Taylor and Laurent Series

A function $f(z)$, which is single-valued and analytic at the point $z=z_{0}$, can be expanded in a power series of Taylor in the neighborhood of this point:

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

the coefficients of which $c_{n}$ are calculated by the formulas

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

where $\Gamma$ is a circle centered at the point $z=z_{0}$, entirely lying in the neighborhood of the point $z_{0}$, where the function $f(z)$ is analytic. The center of the circle of convergence is at the point $z_{0}$; this circle passes through the singular point $\zeta$ of the function $f(z)$, closest to the point $z_{0}$, i.e., the radius of convergence of the series (8) will be equal to the distance from the point $z_{0}$ to the nearest singular point of the function $f(z)$.

For the functions

$$
\ln (1+z), \quad(1+z)^{\alpha}
$$

the following Taylor series expansions in the neighborhood of the point $z_{0 .}=0$ are valid:

$$
\begin{aligned}
\ln (1+z) & =z-\frac{z^{2}}{2}+\frac{z^{3}}{3}-\ldots+(-1)^{n-1} \frac{z^{n}}{n}+\ldots \quad(R=1) \\
(1+z)^{\alpha} & =1+\alpha z+\frac{\alpha(\alpha-1)}{2!} z^{2}+\frac{\alpha(\alpha-1)(\alpha-2)}{3!} z^{3}+\ldots \\
& \cdots+\frac{\alpha(\alpha-1) \ldots(\alpha+n-1)}{n!} z^{n}+\ldots \quad(R=1)
\end{aligned}
$$

In particular, for $\alpha=-1$ we get

$$
\frac{1}{1+z}=1-z+z^{2}-\ldots+(-1)^{n} z^{n}+\ldots \quad(R=1)
$$

Formula (10) gives the Taylor series expansion in the neighborhood of the point $z=0$ of the principal value of the logarithm; to obtain the Taylor series for other values of the multivalued function $\operatorname{Ln}(1+z)$, one should add the numbers $2 n \pi i, n= \pm 1, \pm 2, \ldots$ to the series (10):

$$
\ln (1+z)=z-\frac{z^{2}}{2}+\frac{z^{3}}{3}-\ldots+2 n \pi i
$$