NoteNextra-origin/pages/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.