NoteNextra-origin/pages/Math416/Math416_L24.md

Math416 Lecture 24

Continue on Generalized Cauchy's Theorem

Homotopy

A homotopy between two curves \gamma_0, \gamma_1 : [0, 1] \to \mathbb{C} is a continuous map H : [0, 1] \times [0, 1] \to \mathbb{C} such that H(z, 0) = \gamma_0(z) and H(z, 1) = \gamma_1(z) for all z \in [0, 1].

Lemma:

Let \Omega be open in \mathbb{C}, Let \gamma_0, \gamma_1 be closed contour, homotopic in \Omega. Then \operatorname{ind}_{\gamma_0} (z) = \operatorname{ind}_{\gamma_1} (z) for all z \in \Omega.

Proof:

Let H(s,t) be a homotopy between \gamma_0 and \gamma_1. Let z_0\in \mathbb{C} \setminus \Omega.

Defined \phi:[0,1]\times[0,1]\to \mathbb{C}\setminus \{0\}, \phi(s,t)=H(s,t)-z_0.

By Technical Lemma, \exists continuous \psi:[0,1]\times[0,1]\to \mathbb{C} such that e^{\psi}=\phi.

For each t, \gamma_t(s)=H(s,t) is a closed curve.

\operatorname{ind}_{\gamma_t}(z_0)=\frac{1}{2\pi i}\left[\psi(1,t)-\psi(0,t)\right].

This is continuous (in t), integer valued, thus constant.

QED

Theorem 9.14 Homotopy version of Cauchy's Theorem

Let \Omega be open, \gamma_0, \gamma_1 be two piecewise continuous curves in \Omega that are homotopic.

Then \int_{\gamma_0} f(z) \, dz = \int_{\gamma_1} f(z) \, dz for all f\in O(\Omega).

Proof:

\Gamma=\gamma_0-\gamma_1, then \operatorname{ind}_{\Gamma}(z)=0 for all z\in \mathbb{C}\setminus \Omega.

QED

Corollary of Theorem 9.14

If \gamma_0 is null-homotopic in \Omega (i.e. \gamma_0 is homotopic to a point), then \int_{\gamma_0} f(z) \, dz = 0 for all f\in O(\Omega).

Chapter 10: Further development of Complex Function Theory

Simple connectedness

Definition (non-standard) simply connected

Let \Omega be a domain in \mathbb{C}. We say \Omega is simply connected if \overline{\mathbb{C}}\setminus \Omega is connected. (\overline{\mathbb{C}}=\mathbb{C}\cup \{\infty\})

Example:

disk is simply connected.

annulus is not simply connected.

\mathbb{C} is simply connected.

Any convex domain is simply connected.

Standard definition: \Omega is simply connected if every closed curve in \Omega is null-homotopic in \Omega.

Theorem of equivalent definition of simply connected

For open connected subsets of \mathbb{C}, the standard definition and the non-standard definition are equivalent.

Proved end of book.

Proposition for simply connected domain

\Omega is simply connected \iff every contour in \Omega has winding number 0 about every point in \mathbb{C}\setminus \Omega.

Proof:

If \Omega is simply connected, let \gamma be a curve in \Omega, then \operatorname{ind}_{\gamma}(z)=0 for all z in the unbounded component of \overline{\mathbb{C}}\setminus \Omega. This contains all of \mathbb{C}\setminus \Omega.

Conversely, assume \Omega is not simply connected, then \exists K\cup L=\overline{\mathbb{C}}\setminus \Omega, where K and L are disjoint closed, without loss of generality, assume \infty\in L.

Let H=\Omega\cup K=\mathbb{C}\setminus L.

H is open, K is compact subset of H, so by Separation Lemma, \exists \gamma\in H\setminus K=\Omega such that K\subset \operatorname{int}(\gamma).

Theorem 10.3 Cauchy's Theorem for simply connected domain

corollary of Proposition for simply connected domain

Let \Omega be a simply connected domain, let \gamma be a closed curve in \Omega. Then \int_{\gamma} f(z) \, dz = 0 for all f\in O(\Omega).

Proof:

Know that is true if \operatorname{ind}_{\gamma}(z)=0 for all z\in \mathbb{C}\setminus \Omega.

By Proposition, \Omega is simply connected \iff every closed curve in \Omega has winding number 0 about every point in \mathbb{C}\setminus \Omega.

So the result is true.

QED

Theorem 10.4-6

The following condition are equivalent:

1. \Omega is simply connected.
2. every holomorphic function on \Omega has a primitive g, i.e. g'(z)=f(z) for all z\in \Omega.
3. every non-vanishing holomorphic function on \Omega has a holomorphic logarithm.
4. every harmonic function on \Omega has a harmonic conjugate.

Proof:

(1)\iff (2):

First we show (1)\implies (2).

Assume \Omega is simply connected.

Define g(z)=\int_{z_0}^{z} f(w) \, dw for z_0\in \Omega fixed. Then by Cauchy's Theorem, this definition does not depend on the path.

\frac{g(z+h)-g(z)}{h}=\frac{1}{h}\left[\int_{z}^{z+h} f(w) \, dw\right]

\frac{1}{h}\left[\int_{z}^{z+h} f(w) \, dw\right]\to f(z) as h\to 0.

So on [z,z+h]\subset \Omega, if |f(w)-f(z)|<\epsilon, then |\frac{g(z+h)-g(z)}{h}-hf(z)|<h\epsilon.

To show (2)\implies (1), we prove \neg (1)\implies \neg (2).

(1)\iff (3):

If \Omega is not simply connected, there is some closed curve \gamma and some z_0\not in \Omega such that \operatorname{ind}_{\gamma}(z_0)\neq 0.

SO \int_{\gamma} \frac{1}{z-z_0} \, dz\neq 0.

So \frac{1}{z-z_0} does not have a primitive on \Omega. \frac{1}{z-z_0} have no logarithm on \Omega.

This shows (3)\implies (1).

Suppose \Omega is simply connected.$f\in O(\Omega)$ and f is non-vanishing. We want to show that f has a logarithm on \Omega.

Let z_0\in \Omega be fixed. And a\in \log f(z_0).

\frac{f'(z)}{f(z)}=g' Since g\in O(\Omega), $we can assume g(z_0)=a.

g(z)=a+\int_{z_0}^{z} g'(w) \, dw

So

(fe^{-g})'=f'e^g+fe^g g'=f(e^g)'=0

So fe^{-g}=c for some c\in \mathbb{C}.

So f=ce^g

QED

Continue on Residue Theorem on Thursday.