Example. Expand into Laurent series in powers of $z$ the function

$$
f(z)=\frac{z+2}{z^{2}-2 z-3}
$$
## SOLUTION.

1. The fraction is proper.

We find the roots of the equation $z^{2}-2 z-3=0$. We have two simple roots $z_{1}=-1$ and $z_{2}=3$.

2. The points $z_{1}=-1$ and $z_{2}=3$ are singular points of the function $f(z)$ (at these points $f(z)$ is not analytic).

The rings of analyticity of the function $f(z)$:

$$
\begin{aligned}
& |z|<1 \\
& 1<|z|<3 \\
& |z|>3
\end{aligned}
$$

3. We decompose $f(z)$ into partial fractions:

$$
\frac{z+2}{(z+1)(z-3)}=-\frac{1}{4} \frac{1}{z+1}+\frac{5}{4} \frac{1}{z-3}
$$

4. In each ring of analyticity, we expand the partial fractions into series using Taylor series expansions.

For $|z|<1$:

$$
\frac{1}{z+1}=\sum_{n=0}^{\infty}(-1)^{n} z^{n}, \quad \frac{1}{z-3}=-\frac{1}{3} \sum_{n=0}^{\infty} \frac{z^{n}}{3^{n}}=-\sum_{n=0}^{\infty} \frac{z^{n}}{3^{n+1}}
$$

Thus, in the ring $|z|<1$, the Laurent series of $f(z)$ is:

$$
\frac{z+2}{z^{2}-2 z-3}=-\frac{1}{4} \sum_{n=0}^{\infty}(-1)^{n} z^{n}+\frac{5}{4} \sum_{n=0}^{\infty} \frac{z^{n}}{3^{n+1}}=\frac{1}{4} \sum_{n=0}^{\infty}\left[(-1)^{n+1}-\frac{5}{3^{n+1}}\right] z^{n}
$$

For $1<|z|<3$:

$$
\frac{1}{z+1}=-\frac{1}{z} \frac{1}{1-\frac{1}{z}}=-\sum_{n=0}^{\infty} \frac{1}{z^{n+1}}=-\sum_{n=1}^{\infty} \frac{1}{z^{n}}, \quad |z|>1
$$

$$
\frac{1}{z-3}=\frac{1}{z} \frac{1}{1-\frac{3}{z}}=\sum_{n=0}^{\infty} \frac{3^{n}}{z^{n+1}}=\sum_{n=1}^{\infty} \frac{3^{n-1}}{z^{n}}, \quad |z|>3
$$

Thus, in the ring $1<|z|<3$, the Laurent series of $f(z)$ is:

$$
\frac{z+2}{z^{2}-2 z-3}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{4 \cdot z^{n}}-\sum_{n=0}^{\infty} \frac{5 \cdot z^{n}}{4 \cdot 3^{n+1}}
$$

For $|z|>3$:

The series (3) converges, and the series (2) diverges. Therefore, instead of (2), we use:

$$
\frac{1}{z-3}=\frac{1}{z} \frac{1}{1-\frac{3}{z}}=-\sum_{n=0}^{\infty} \frac{3^{n}}{z^{n+1}}=-\sum_{n=1}^{\infty} \frac{3^{n-1}}{z^{n}}, \quad|z|>3
$$

Thus, in the region $|z|>3$, the Laurent series of $f(z)$ is:

$$
\frac{z+2}{z^{2}-2 z-3}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{4 \cdot z^{n}}-\frac{5}{4} \sum_{n=1}^{\infty} \frac{3^{n-1}}{z^{n}}=\sum_{n=1}^{\infty} \frac{(-1)^{n}+5 \cdot 3^{n-1}}{4} \frac{1}{z^{n}}
$$

Answer.

$$
\begin{aligned}
& \frac{z+2}{z^{2}-2 z-3}=\frac{1}{4} \sum_{n=0}^{\infty}\left[(-1)^{n+1}-\frac{5}{3^{n+1}}\right] z^{n}, \quad|z|<1 \\
& \frac{z+2}{z^{2}-2 z-3}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{4 \cdot z^{n}}-\sum_{n=0}^{\infty} \frac{5 \cdot z^{n}}{4 \cdot 3^{n+1}}, \quad 1<|z|<3 \\
& \frac{z+2}{z^{2}-2 z-3}=\sum_{n=1}^{\infty} \frac{(-1)^{n}+5 \cdot 3^{n-1}}{4} \frac{1}{z^{n}}, \quad 3<|z|<\infty
\end{aligned}
$$

Conditions of the problems. Expand the functions $f(z)$ into Laurent series in powers of $z$.

1. $f(z)=1 /(z-2)$. 2. $f(z)=1 /\left(z^{2}+z\right)$. 3. $f(z)=2 /\left(z^{2}+2 z\right)$. 4. $f(z)=1 /\left(z^{4}-z^{2}\right)$. 5. $f(z)=1 /\left(z^{3}-3 z^{2}\right)$. 6. $f(z)=(2 z-3) /\left(z^{2}-5 z+6\right)$. 7. $f(z)=2 /\left(z^{3}-4 z^{2}+3 z\right)$. 8. $f(z)=1 /\left(z^{4}-3 z^{3}+2 z^{2}\right)$. 9. $f(z)=$ $=(2 z+3) /\left(z^{3}+3 z^{2}+2 z\right)$. 10. $f(z)=3 /\left(z^{4}+5 z^{2}+4\right)$.

## Answers.

1. $f(z)=-\sum_{n=0}^{\infty} \frac{z^{n}}{2^{n+1}}, 0<|z|<2, \quad f(z)=\sum_{n=0}^{\infty} \frac{2^{n}}{z^{n+1}}, 2<|z|<\infty$.
2. $f(z)=\sum_{n=0}^{\infty}(-1)^{n} z^{n-1}, 0<|z|<1, \quad f(z)=\sum_{n=0}^{\infty} \frac{1}{z^{n+2}}, \quad 1<|z|<\infty$.
3. $f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n} z^{n-1}}{2^{n}}, 0<|z|<2, f(z)=\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{n+1}}{z^{n+2}}, 2<|z|<\infty$.
4. $f(z)=-\sum_{n=0}^{\infty} z^{2 n-2}, 0<|z|<1, \quad f(z)=\sum_{n=0}^{\infty} \frac{1}{z^{2 n+4}}, \quad 1<|z|<\infty$.
5. $f(z)=-\sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}}, 0<|z|<3, \quad f(z)=\sum_{n=0}^{\infty} \frac{3^{n}}{z^{n+3}}, \quad 3<|z|<\infty$.
6. $f(z)=-\sum_{n \propto 0}^{\infty}\left(\frac{1}{2^{n+1}}+\frac{1}{3^{n+1}}\right) z^{n}, \quad|z|<2$,

$$
\begin{aligned}
& f(z)=-\sum_{x=0}^{n_{\infty} \mid 0}\left(\frac{2^{n}}{z^{n+1}}-\frac{z^{n}}{3^{n+1}}\right), \quad 2<|z|<3
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_718fd8b5b0aaabbbc775g-040.jpg?height=101&width=552&top_left_y=1194&top_left_x=197)
7. $f(z)=\sum_{n=0_{\infty}}^{\infty}\left(1-\frac{1}{3^{n}}