NoteNextra-origin/content/Math416/Math416_L20.md

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