Example. Expand the function in a Laurent series in the neighborhood of its singular point

$$
f(z)=z \sin \frac{\pi z}{z-1}
$$

Find the region where the obtained expansion is valid.
Solution. The function $f(z)$ has a unique singular point $z=1$, therefore, it must be expanded into a Laurent series in powers of $z-1$.

1. We introduce an auxiliary variable $t=z-1$. We obtain

$$
f(t+1)=(t+1) \sin \frac{\pi t+1}{t}
$$

2. We transform the function $f(t+1)$ into a form that allows the use of tabulated expansions:

$$
f(t+1)=(t+1) \sin \left(\pi+\frac{1}{t}\right)=-t \sin \frac{1}{t}-\sin \frac{1}{t} .
$$

3. Using the tabulated Taylor series expansion

$$
\sin z=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} z^{2 n+1}, \quad|z|<\infty
$$

and substituting $z$ with $1 / t$, we find the Laurent series expansion of $\sin (1 / t)$

$$
\sin \frac{1}{t}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{t^{2 n+1}}, \quad 0<|t|<\infty
$$

Substituting into (1), we get

$$
f(t)=-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{t^{2 n}}-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{t^{2 n+1}}
$$

4. Replacing $t$ with $z-1$, we obtain

$$
f(z)=-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{(z-1)^{2 n}}-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{(z-1)^{2 n+1}}
$$

5. We find the domain in which the obtained expansion is valid. Since $f(z)$ has no other singular points except $z=1$, the obtained Laurent series converges to $f(z)$ for all $z \neq 1$.

Answer. $f(z)=$

$=-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{(z-1)^{2 n}}-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!} \frac{1}{(z-1)^{2 n+1}}, 0<|z-1|<\infty$.

Problem Statements. Expand the function $f(z)$ into a Laurent series in the neighborhood of its singular point.

1. $f(z)=\sin (1 / z)$. 2. $f(z)=e^{-1 / z^{2}}$. 3. $f(z)=z \cos (1 / z)$. 4. $f(z)=$ $=e^{z /(z-1)} . \quad$ 5. $f(z)=\cos (\pi z /(z+1)) .6 . f(z)=\sin [(z+2) /(z+1)]$. 7. $f(z)=e^{z /(z-i)}$. 8. $f(z)=z \cos [\pi z /(z+2)]$. 9. $f(z)=z e^{z /(z+3)}$. 10. $f(z)=1 /\left(1-z^{2}\right)$.

## Answers.

1. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!z^{2 n+1}}, 0<|z|<\infty$.
2. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!z^{2 n}}, 0<|z|<\infty$.
3. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n)!z^{2 n-1}}, 0<|z|<\infty$.
4. $f(z)=e \sum_{n=0}^{\infty} \frac{1}{n!(z-1)^{n}}, 0<|z-1|<\infty$.
5. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \pi^{2 n}}{(2 n)!(z+1)^{2 n}}, 0<|z+1|<\infty$.
6. $f(z)=\sin 1 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n)!(z+1)^{2 n}}+\cos 1 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)!(z+1)^{2 n+1}}$, $0<|z+1|<\infty$.
7. $f(z)=e \sum_{n=0}^{\infty} \frac{i^{n}}{n!(z-i)^{n}}, 0<|z-i|<\infty$.
8. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2 n}}{(2 n)!(z+2)^{2 n-1}}-\sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2^{2 n+1}}{(2 n)!(z+2)^{2 n}}, \quad 0<|z+2|<\infty$.
9. $f(z)=e\left[\sum_{n=0}^{\infty} \frac{(-1)^{n} 3^{n}}{n!(z+3)^{n-1}}-\sum_{n=0}^{\infty} \frac{(-1)^{n} 3^{n+1}}{n!(z+3)^{n}}\right], 0<|z+3|<\infty$.
10. $f(z)=\sum_{n=0}^{\infty} \frac{(z+1)^{n-1}}{2^{n+1}}, 0<|z+1|<2$,

$$
f(z)=-\sum_{n=0}^{\infty} \frac{(z-1)^{n-1}}{2^{n+1}}, 0<|z-1|<2
$$

### 1.12. Zeros of an Analytic Function

Problem Statement. Find the zeros of the analytic function $f(z)$ and determine their order (multiplicity).

Solution Plan.

1. Find the zeros of the analytic function $f(z)$ by solving the equation $f(z)=0$.
2. Determine the order of each obtained zero $z=z_{0}$. For this, use one of the following (equivalent) statements:

a) if the power series expansion (Taylor series) of $f(z)$ in the neighborhood of the zero $z=z_{0}$ has the form

$$
f(z)=c_{n}\left(z-z_{0}\right)^{n}+c_{n+1}\left(z-z_{0}\right)^{n+1}+\ldots, \quad \text { where } c_{n} \neq 0
$$

then $z_{0}$ is a zero of order $n$ of the function $f(z)$;

b) if $f\left(z_{0}\right)=f^{\prime}\left(z_{0}\right)=\ldots=f^{n-1}\left(z_{0}\right)=0$ and $f^{n}\left(z_{0}\right) \neq 0$, then $z_{0}$ is a zero of order $n$ of the function $f(z)$

c) if the function $f(z)$ can be represented as

$$
f(z)=\left(z-z_{0}\right)^{n} \varphi(z)
$$

where $\varphi(z)$ is analytic at $z_{0}$ and $\varphi\left(z_{0}\right) \neq 0$, then $z_{0}$ is a zero of order $n$ of the function $f(z)$.

Remark. If the function $f(z)=f_{1}(z) f_{2}(z)$, and $z_{0}$ is a zero of order $n_{1}$ of the function $f_{1}(z)$ and $z_{0}$ is a zero of order $n_{2}$ of the function $f_{2}(z)$, then $z_{0}$ is a zero of order $n_{1}+n_{2}$ of the function $f(z)$.