# Math416 Lecture 19

## Continue on the Laurent series

### Laurent series

If $f$ is holomorphic in $A(z_0;R_1,R_2)$ then $f=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ where the Laurent series converges on the annulus $A(z_0;R_1,R_2)$

$$
\int_{C(z_0,r)} f(z)(z-z_0)^{-k-1} dz = \sum_{n=-\infty}^{\infty} a_n \int_{C(z_0,r)} (z-z_0)^{n-k-1} dz=a_k 2\pi i
$$

> $C(z_0,r)$ is a circle centered at $z_0$ with radius $r$

### Isolated singularities

A punctured disk at $z_0$ is $A(z_0;0,R)=\{z:0<|z-z_0|<R\}$

Say a function $f$ has an isolated singularity at $z_0$ if it is holomorphic in a punctured disk $A(z_0;0,R)$

$f$ has a Laurent series in $A(z_0;0,R)$

$$
f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n
$$

that converges in $A(z_0;0,R)$

#### Principal part of a Laurent series

The principal part of a Laurent series is the sum of the terms with negative powers of $(z-z_0)$

$$
\sum_{n=-\infty}^{-1} a_n (z-z_0)^n
$$

Say the isolated singularity is

- removable if $a_n=0$ for all $n<0$
  - If $f(z)$ has a removable singularity at $z_0$, then extend $f$ to $\mathbb{D}_{z_0,R}$ by defining $f(z_0)=a_0$. This extended $f$ is holomorphic on $\mathbb{D}_{z_0,R}$ and $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$ for $z\in \mathbb{D}_{z_0,R}$
- pole if $a_{-k}\neq 0$ and $a_n=0$ for all $n<-k$
  - A pole with order $1$ is a simple pole
- essential if the cases above are not true

Example:

1. $f(z)=\frac{\sin z}{z}$ has a removable singularity at $z=0$.

the power series is

$$
\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots
$$

So the Laurent series is

$$
\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots
$$

The singularity is removable by defining $f(0)=1$

2. $f(z)=\frac{z^2-1}{(z-1)(z-3)}=\frac{(z-1)(z+1)}{(z-1)(z-3)}$

There are two poles at $z=1$ and $z=3$

the singularity at $z=1$ is removable by defining $f(1)=1$

the singularity at $z=3$ is a simple pole with order 1 $f(z)=\frac{z+1}{z-3}=\frac{(z-3)+4}{z-3}=4(z-3)^{-1}+1$

3. $f(z)=\frac{(z+1)^2(z+2)^3}{(z-1)^2(z-5)^6(z-8)}$

there are three poles at $z=1,5,8$, the order of the poles are 2, 6, 1 respectively.

#### Corollary: order of poles and zeros

If $f$ has a pole of order $m$ at $z_0$,

$$
f(z) = \sum_{n=-m}^{\infty} a_n (z-z_0)^n
$$

then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$. Value of holomorphic extension of $(z-z_0)^m f(z)$ at $z_0$ is $a_{-m}$.

- $f$ is given by a power series in $A(z_0;0,R)$
- $f=(z-z_0)^{-m} g(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$, $\frac{1}{f}=(z-z_0)^m \frac{1}{g(z)}$ has a pole of order $m$ at $z_0$. So $f$ has a pole of order $m$ at $z_0$ if and only if $\frac{1}{f}$ has a zero of order $m$ at $z_0$

> $e^{1/z}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ has an essential singularity at $z=0$ since it has infinitely many terms with negative powers of $z$.

Suppose $f$ is a holomorphic in a neighborhood of $\infty$: $\exists R>0$ s.t. $f$ is holomorphic on $\{z:|z|>R\}$

We defined $g(z)=f(1/z)$ where $g$ is holomorphic on punctured disk center $0$ radius $1/R$

Say $f(z)$ has a zero of order $\infty$ if any only if $g(z)=f(1/z)$ has a zero of order $m$ at $z=0$

Say $f$ has a pole of order $m$ at $\infty$ if and only if $g(z)=f(1/z)$ has a pole of order $m$ at $z=0$

Example:

1. $f(z)=z^2$, $g(z)=f(1/z)=1/z^2$ has a pole of order 2 at $z=0$
2. $f(z)=\frac{1}{z^3}$ (vanishes to order 3 at $\infty$), $g(z)=f(1/z)=z^3$ has a zero of order 3 at $z=0$

We say $f$ has an isolated singularity at $\infty$ if and only if $g(z)=f(1/z)$ has an isolated singularity at $z=0$.

$f$ has $\begin{cases}
  \text{removable}\\
  \text{pole of order } m\\
  \text{essential}
\end{cases}$ singularity at $\infty$ if and only if $g(z)=f(1/z)$ has $\begin{cases}
  \text{removable}\\
  \text{pole of order } m\\
  \text{essential}
\end{cases}$ singularity at $z=0$

#### Theorem: Criterion for a removable singularity (Riemann removable singularity theorem)

Suppose $f$ has an isolated singularity at $z_0$. Then it is removable if and only if $f$ is bounded on a punctured disk centered at $z_0$.

Proof:

($\Leftarrow$) Suppose $z_0$ is a removable singularity. Then $\exists r>0$ such that $B_r(z_0)\setminus\{z_0\}=A(z_0;0,r)$ and $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$ for $z\in A(z_0;0,r)$. Then $f$ is bounded in $A(z_0;0,r/2)$

($\Rightarrow$) Suppose $|f(z)|\leq M$ for $z\in A(z_0;0,r/2)$. So $f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^{n-k-1}=\int_{C_r}f(z)(z-z_0)^{-k-1}dz=a_{k}2\pi i$

$a_k=\frac{1}{2\pi i}\int_{C_r}f(z)(z-z_0)^{-k-1}dz$

And $|a_k|\leq \max_{z\in C_r}\left|2\pi|f(z)|z-z_0|^{-k-1}\right|\leq 2\pi M r^{-k-1}2\pi r$

So $|a_k|\leq (4\pi^2M)r^{-k}$ for all $r<R$

So if $k<0$, $|a_k|\leq \lim_{r\to 0} (4\pi^2M)r^{-k}=0$

QED

Corollary:

If $f$ is holomorphic at $\infty$, then $f$ is bounded for large $|z|$.

#### Theorem: Criterion for a pole

Suppose $f$ has an isolated singularity at $z_0$. Then $z_0$ is a pole of order $m$ if and only if $\lim_{z\to z_0} |f(z)|=\infty$

Proof:

($\Rightarrow$) If $z_0$ is a pole of order $m$, then $f(z)=a_{-m}(z-z_0)^{-m}+O((z-z_0)^{-m+1})$

As $z\to z_0$, $|f(z)|\approx |a_{-m}| |z-z_0|^{-m}\to \infty$

($\Leftarrow$) Let $g(z)=\frac{1}{f(z)}$ near $z_0$. Then $g$ has a singularity at $z_0$ and $|g(z)|$ is bounded near $z_0$.

By Riemann removable singularity theorem, $g(z)=(z-z_0)^m h(z)$ for some holomorphic $h$ and $h(z_0)\neq 0$

So $f(z)=\frac{1}{g(z)}=\frac{1}{(z-z_0)^m h(z)}$ has a pole of order $m$ at $z_0$

QED