NoteNextra-origin/content/Math416/Math416_L19.md

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