# Math416 Lecture 18

## Chapter 8: Laurent Series and Isolated Singularities

### 8.1 Laurent Series

#### Definition of Laurent Series

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

where $c_n$ are complex coefficients.

Let $R_2=\frac{1}{\limsup_{n\to\infty} |c_n|^{1/n}}$, then the Laurent series converges on $|z-z_0|<R_2$

Where $R_1=\limsup_{n\to-\infty} |c_n|^{-1/n}$, if $|z-z_0|>R_1$, the Laurent series diverges.

If $R_1\leq R_2$, then the Laurent series converges on $A(z_0;R_1,R_2)=\{z:R_1<|z-z_0|<R_2\}$, the Laurent series converges absolutely on $A(z_0;R_1,R_2)$

By Weierstrass, the limit is a holomorphic function on $A(z_0;R_1,R_2)$

If $R_1<r<R_2$, then

$$
\int_{C(z_0,r)} f(z) dz = \sum_{n=-\infty}^{\infty} c_n \int_{C(z_0,r)} (z-z_0)^n dz
$$

<details>
<summary>Additional Proof</summary>

$$
\int_{C(z_0,r)} (z-z_0)^n dz = \begin{cases} 2\pi i, & n=-1 \\0, & n\neq -1\end{cases}
$$

Proof:

$\gamma(t)=z_0+re^{it}, t\in[0,2\pi]$
$$
\begin{aligned}
\int_{C(z_0,r)} (z-z_0)^n dz &= \int_0^{2\pi} (z_0+re^{it}-z_0)^n ire^{it} dt \\
&= ir^{n+1} \int_0^{2\pi} e^{i(n+1)t} dt \\
&= \begin{cases}
    2\pi i, & n=-1 \\
    \int_0^{2\pi} e^{i(n+1)t} dt = \frac{1}{i(n+1)}e^{i(n+1)t}\Big|_0^{2\pi} = 0, & n\neq -1
\end{cases}
\end{aligned}
$$
</details>

So,

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

And,

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

So,

$$
2\pi i c_j = \int_{C(z_0,r)} f(z)(z-z_0)^{-j-1} dz
$$

### Cauchy integral

Recall Cauchy integral formula:

$$
f(z) = \int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi
$$

where $\gamma$ is a closed curve.

Suppose $|z-z_0|>R$,

$$
\begin{aligned}
\frac{1}{\xi-z}&=\frac{1}{\xi-z_0-(z-z_0)}\\
&=-\frac{1}{z-z_0}\frac{1}{1-\frac{\xi-z_0}{z-z_0}}\\
&=-\frac{1}{z-z_0}\sum_{n=0}^{\infty} \frac{(\xi-z_0)^n}{(z-z_0)^n}\\
&=-\sum_{m=1}^{\infty} (\xi-z_0)^{m-1}(z-z_0)^{-m}
\end{aligned}
$$

So,

$$
\begin{aligned}
f(z) &= \int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi\\
&= -\int_{\gamma} \sum_{m=1}^{\infty} (\xi-z_0)^{m-1}(z-z_0)^{-m}\phi(\xi) d\xi\\
&=-\sum_{m=1}^{\infty} (z-z_0)^{-m} \int_{\gamma} (\xi-z_0)^{m-1} \phi(\xi) d\xi
\end{aligned}
$$

So the Cauchy integral $\int_{\gamma} \frac{\phi(\xi)}{\xi-z} d\xi$ is a convergent power series in $B_{d(z_0,\gamma)}(z_0)$ and is a convergent Laurent series (with just negative powers) in $\mathbb{C}\setminus B_{\max_{\xi\in \gamma} d(z_0,\xi)}(z_0)$

#### Theorem 8.4 Cauchy Theorem for Annulus

Suppose $f$ is holomorphic on $A(z_0;R_1,R_2)$, Let $C_r=\{z:|z-z_0|=r\}$, oriented counterclockwise. Then $I(r)=\int_{C_r} f(z) dz$ is independent of $r$ for $R_1<r<R_2$

Proof:

If integrand is continuous with respect to $r$ and continuous with respect to $t$, then we can differentiate under the integral sign (Check after class, on Appendix 4?)

$$
\begin{aligned}
I(r)&=\int_0^{2\pi} f(z_0+re^{it})ire^{it}dt\\
\frac{dI}{dr}&=i\int_0^{2\pi}\frac{\partial}{\partial r}[f(z_0+re^{it})re^{it}]dt
\end{aligned}
$$

$$
\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]=if'(z_0+re^{it})e^{it}+if(z_0+re^{it})e^{it}
$$

$$
\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]=f'(z_0+re^{it})ire^{it}+f(z_0+re^{it})ire^{it}
$$

This gives

$$
\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]=\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]
$$

So,

$$
\frac{dI}{dr}=i\int_0^{2\pi}\frac{\partial}{\partial r}[f(z_0+re^{it})ire^{it}]dt=i\int_0^{2\pi}\frac{\partial}{\partial t}[f(z_0+re^{it})e^{it}]dt=0
$$

is a integration on a closed curve, so it is $0$.

So, $I(r)$ is constant.

QED

Let $f$ be holomorphic on $A(z_0;R_1,R_2)$. Let $C_r=\{z:|z-z_0|=r\}$, oriented counterclockwise. Let $w\in A(z_0;R_1,R_2)$. Choose $R_1<r_1<|w-z_0|<r_2<R_2$ such that $w\in A(z_0;r_1,r_2)$. Then,

$$
f(w)=\frac{1}{2\pi i}\int_{C_{r_2}} \frac{f(z)}{z-w} dz-\frac{1}{2\pi i}\int_{C_{r_1}} \frac{f(z)}{z-w} dz
$$

Proof:

Define $g(z)=\begin{cases}
    \frac{f(z)-f(w)}{z-w}, & z\neq w \\
    f'(w), & z=w
\end{cases}$

Then $g$ is holomorphic on $A(z_0;r_1,r_2)$ since $f$ is analytic at $w$, $f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(w)}{n!}(z-w)^n$

So,

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

So,

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

So,

$$
\lim_{z\to w} \frac{f(z)-f(w)}{z-w}=f'(w)
$$

So $\int_{C_{r_2}} g(z) dz=\int_{C_{r_1}} g(z) dz$,

$$
\int_{C_{r_2}} \frac{f(z)}{z-w} dz-f(w)\int_{C_{r_2}} \frac{1}{z-w} dz=\int_{C_{r_1}} \frac{f(z)}{z-w} dz-f(w)\int_{C_{r_1}} \frac{1}{z-w} dz
$$

Since $\int_{C_{r_2}} \frac{1}{z-w} dz=2\pi i$ and $\int_{C_{r_1}} \frac{1}{z-w} dz=0$, (using Cauchy integral theorem on convex region)

$$
f(w)=\frac{1}{2\pi i}\int_{C_{r_2}} \frac{f(z)}{z-w} dz-\frac{1}{2\pi i}\int_{C_{r_1}} \frac{f(z)}{z-w} dz
$$

QED

Since $\int_{C_{r_1}} \frac{f(z)}{z-w} dz$ is a Laurent series in negative powers which converges in $\mathbb{C}\setminus \overline{B_{r_1}(z_0)}$, we can conclude that

$f(z)$ is given by a convergent Laurent series $\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ in $\mathbb{C}\setminus \overline{B_{r_1}(z_0)}$ where $a_n=\frac{1}{2\pi i}\int_{C_r} \frac{f(z)}{(z-z_0)^{-1-n}} dz$

Laurent series converges in $A(z_0;R_1,R_2)$