# Laurent series expansion (example)

$$\text{Let} \quad f(z)=\frac{1}{z}+\frac{2}{3-z},\quad z\in \mathbb{C}\smallsetminus \{0,3\}.$$ To find the Laurent series expansion of $f$ about $z_0=1$ within the annulus $1<|z-1|<2,$ one must aim to expand each fraction as an infinte geometric series with radius of convergence $|r|<1.$ $$f(z)=\frac{1}{1+(z-1)}+\frac{2}{2-(z-1)}=\frac{1}{(z-1)} \cdot \frac{1}{1+\frac{1}{z-1}}+\frac{1}{1-\frac{z-1}{2}}$$ Since $\bigg|\frac{z-1}{2}\bigg|<1$ and $\bigg|\frac{1}{z-1}\bigg|<1,$ $$f(z)=\frac{1}{(z-1)} \cdot\sum_{n=0}^\infty \frac{(-1)^n}{(z-1)^n}+\sum_{n=0}^\infty \frac{(z-1)^n}{2^n} \iff$$ $$f(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(z-1)^{n+1}}+\sum_{n=0}^\infty \frac{(z-1)^n}{2^n}$$ Can someone elaborate on the geometric relation between the large and small radius of the annulus and the geometric series' radii of convergence?

## 2 Answers

As you see the series $$f(z)=\frac{1}{(z-1)} \cdot\sum_{n=0}^\infty \frac{(-1)^n}{(z-1)^n}$$ valid for $\bigg|\frac{1}{z-1}\bigg|<1$ which is outside a circle centered at $z=1$, and the series $$\sum_{n=0}^\infty \frac{(z-1)^n}{2^n}$$ valid for $\bigg|\frac{z-1}{2}\bigg|<1$, which is inside of a circle centered at $z=1$, so your Laurent series $$f(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(z-1)^{n+1}}+\sum_{n=0}^\infty \frac{(z-1)^n}{2^n}$$ valid in intersection of two areas, that is $$\{z:|z-1|<2\} \bigcap \{z:|z-1|>1\} = \{z:1<|z-1|<2\}$$

The function \begin{align*} f(z)=\frac{1}{z}+\frac{2}{3-z}\qquad\qquad z\in\mathbb{C}\setminus\{0,3\} \end{align*} is to expand around the center $z_0=1$.

Informally: In order to determine the regions of convergence in case of isolated singularities we start from the point we want to expand the function (here $z_0=1$). Then we look at the greatest circular region with center $z_0$ which does not contain a singularity and so has necessarily a point of singularity at its boundary. This determines the first region. Then we consider an annulus starting from the first point of singularity circularly extended to the next point of singularity. We continue this way till all singularities are at the boundary of one of these regions. The final region is the complement of (the union of the closure) of all the other regions.

Since there are simple poles at $z=0$ and $z=3$ we have to distinguish three regions of convergence \begin{align*} D_1:&\quad 0\leq |z-1|<1\\ D_2:&\quad 1<|z-1|<3\\ D_3:&\quad |z-1|>3 \end{align*}

• The first region $D_1$ is a disc with center $z=1$, radius $1$ and the pole at $z=0$ at the boundary of the disc. It admits for both fractions of $f$ a representation as power series.

• The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It has the poles $z=0$ and $z=3$ at the boundary of the annulus. It admits for the fraction with pole at $z=0$ a representation as principal part of a Laurent series and for the fraction with pole at $z=1$ a representation as power series.

• The region $D_3$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.

Hint: An answer with calculation of a power series and principal part is for instance given here.