# Math416 Lecture 20

## Laurent Series and Isolated Singularities

### Isolated Singularities

$f$ has an isolated singularity at $z_0$ if $f$ is analytic everywhere in some punctured disk $0 < |z - z_0| < R$ except at $z_0$ itself.

#### Removable Singularities

We call $z_0$ a removable singularity if there exists $g\in O(B_r(z_0))$ such that $f(z) = g(z)$ for all $z\in B_r(z_0) \setminus \{z_0\}$.

#### Poles

We call $z_0$ a pole if there are finitely many terms with negative powers in the Laurent series expansion of $f$ about $z_0$.

#### Essential Singularities

We call $z_0$ an essential singularity if there are infinitely many terms with negative powers in the Laurent series expansion of $f$ about $z_0$.

#### 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 $z_0$.

#### Theorem 8.10 (Casorati-Weierstrass Theorem)

If $z_0$ is an essential singularity of $f$, then $\forall r>0$, $\overline{f(B_r(z_0)\setminus\{z_0\})}= \mathbb{C}$.

Proof:

Suppose $w\notin$ closure range fo $f$ on $B_r(z_0)\setminus\{z_0\}$, then $\exists \epsilon > 0$ such that $B_\epsilon(w)\cap f(B_r(z_0)\setminus\{z_0\})=\emptyset$.

$g(z)=1/(f(z)-w)$ and $|g(z)|\leq \frac{1}{\epsilon}$, which is bounded. By Riemann removable singularity theorem, $g$ has a removable singularity. So $f(z)=\frac{1}{g(z)}+w$ is holomorphic on $B_r(z_0)\setminus\{z_0\}$.

Suppose $g(z_0)\neq 0$, then $f$ has a removable singularity at $z_0$.

Suppose $g(z_0)=0$, then $f$ has a pole at $z_0$.

This contradicts the assumption that $z_0$ is an essential singularity.

QED

#### Theorem 8.11 (Picard's Theorem)

If $z_0$ is an essential singularity of $f$, then $\forall r>0$, $f(B_r(z_0)\setminus\{z_0\})$ contains every point in $\mathbb{C}$ except possibly one.

#### Definition: Residue

Suppose $f$ has an isolated singularity at $z_0$. The residue of $f$ at $z_0$, write $res_{z_0}(f)$, is the coefficient of $(z-z_0)^{-1}$ in the Laurent series expansion of $f$ about $z_0$.

> Preview:
>
> Residue Theorem:
>
> Suppose $G$ is a simply connected domain, $F$ is a finite set in $G$ $h$ is holomorphic on $G\setminus F$. Let $\gamma$ be a simple closed curve in $G$, containing $\lambda_1, \lambda_2, \cdots, \lambda_N$ from $F$. Then
>
> $$\int_\gamma h(z) dz = 2\pi i \sum_{j=1}^N res_{\lambda_j}(h)$$

Special case:

When $\gamma=\partial B_r(z_0)$, $f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$ converges on $A(z_0;0,R)$, then

$$
\int_{\gamma} f(z) dz = 2\pi i \sum_{n=-\infty}^\infty a_n \int_{\gamma} (z-z_0)^n dz=2\pi i a_{-1}
$$

Example:

1. Find residue of $f(z)=\frac{\sin z}{z^4}$ at $z=0$.

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

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

$res_{z=0}(\frac{\sin z}{z^4}) = \frac{1}{3!}$

2. Find residue of $f(z)=\frac{1}{(z+2)(z-5)}$ at $z=5$ and $z=-2$.

$res_{z=5}(f(z))=(z-5)^{-1}\cdot\frac{1}{z+2}|_{z=5}=\frac{1}{7}$

$res_{z=-2}(f(z))=(z+2)^{-1}\cdot\frac{1}{z-5}|_{z=-2}=-\frac{1}{7}$

#### Corollary of residue

Suppose $f$ has an simple pole at $z_0$. Then $res_{z_0}(f)=\lim_{z\to z_0}(z-z_0)f(z)$.

Proof:

$f(z)=a_{-1}(z-z_0)^{-1}+\sum_{n=0}^\infty a_n(z-z_0)^n$, $(z-z_0)f(z)=a_{-1}+\sum_{n=0}^\infty a_n(z-z_0)^{n+1}$

$\lim_{z\to z_0}(z-z_0)f(z)=\lim_{z\to z_0}(z-z_0)\cdot a_{-1}+\lim_{z\to z_0}(z-z_0)\cdot\sum_{n=0}^\infty a_n(z-z_0)^n=a_{-1}$

QED

#### Find residue for poles with higher order

Suppose $f$ has a pole of order 2 at $z_0$. Then $f(z)=\frac{a_{-2}}{(z-z_0)^2}+\frac{a_{-1}}{z-z_0}+\sum_{n=0}^\infty a_n(z-z_0)^n$, $(z-z_0)^2f(z)=a_{-2}+(z-z_0)a_{-1}+\sum_{n=0}^\infty a_n(z-z_0)^{n+2}$

Method 1:

$res_{z_0}(f)=a_{-1}=\lim_{z\to z_0}\frac{f(z)-\lim_{z\to z_0}(z-z_0)^2f(z)}{(z-z_0)}$

Method 2:

$res_{z_0}(f)=\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)|_{z=z_0}$

So suppose $f$ has a pole of order $n$ at $z_0$. Then $res_{z_0}(f)=\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)|_{z=z_0}$

Proof:

$f(z)=\frac{a_{-n}}{(z-z_0)^n}+\frac{a_{-n+1}}{(z-z_0)^{n-1}}+\cdots+\frac{a_{-1}}{z-z_0}+\sum_{m=0}^\infty a_m(z-z_0)^m$

$(z-z_0)^nf(z)=a_{-n}+(z-z_0)a_{-n+1}+\cdots+(z-z_0)^{n-1}a_{-1}+\sum_{m=0}^\infty a_m(z-z_0)^{m+n}$

$\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=(n-1)!a_{-1}+\sum_{m=0}^\infty a_m(m+n)(m+n-1)\cdots(m+1)(z-z_0)^{m-1}$

$\lim_{z\to z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=a_{-1}$

QED

## Chapter 9: Generalized Cauchy's Theorem

### Simple connectedness

#### Proposition 9.1

Let $\phi$ be a continuous nowhere vanishing function from $[a,b]\subset\mathbb{R}$ to $\mathbb{C}\setminus\{0\}$. Then there exists a continuous function $\psi:[a,b]\to\mathbb{C}$ such that $e^{\psi(t)}=\phi(t)$ for all $t\in[a,b]$.

Moreover, $\psi$ is uniquely determined up to an additive integer multiple of $2\pi i \mathbb{Z}$.

Proof:

Uniqueness:

Suppose $\phi_1$ and $\phi_2$ are both continuous functions so that $e^{\phi_1(t)}=\phi(t)=e^{\phi_2(t)}$ for all $t\in[a,b]$.

Then $e^{\phi_1(t)-\phi_2(t)}=1$ for all $t\in[a,b]$. So $\phi_1(t)-\phi_2(t)=2k\pi i$ for some $k\in\mathbb{Z}$.

Existence:

Continue on Thursday.

QED