160 lines
5.5 KiB
Markdown
160 lines
5.5 KiB
Markdown
# Math416 Lecture 24
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## Continue on Generalized Cauchy's Theorem
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### Homotopy
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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 $z \in [0, 1]$.
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#### Lemma:
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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 $z \in \Omega$.
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Proof:
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Let $H(s,t)$ be a homotopy between $\gamma_0$ and $\gamma_1$. Let $z_0\in \mathbb{C} \setminus \Omega$.
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Defined $\phi:[0,1]\times[0,1]\to \mathbb{C}\setminus \{0\}$, $\phi(s,t)=H(s,t)-z_0$.
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By [Technical Lemma](https://notenextra.trance-0.com/Math416/Math416_L23#lemma-912-technical-lemma), $\exists$ continuous $\psi:[0,1]\times[0,1]\to \mathbb{C}$ such that $e^{\psi}=\phi$.
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For each $t$, $\gamma_t(s)=H(s,t)$ is a closed curve.
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$\operatorname{ind}_{\gamma_t}(z_0)=\frac{1}{2\pi i}\left[\psi(1,t)-\psi(0,t)\right]$.
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This is continuous (in $t$), integer valued, thus constant.
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QED
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#### Theorem 9.14 Homotopy version of Cauchy's Theorem
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Let $\Omega$ be open, $\gamma_0, \gamma_1$ be two piecewise continuous curves in $\Omega$ that are homotopic.
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Then $\int_{\gamma_0} f(z) \, dz = \int_{\gamma_1} f(z) \, dz$ for all $f\in O(\Omega)$.
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Proof:
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$\Gamma=\gamma_0-\gamma_1$, then $\operatorname{ind}_{\Gamma}(z)=0$ for all $z\in \mathbb{C}\setminus \Omega$.
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QED
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#### Corollary of Theorem 9.14
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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 $f\in O(\Omega)$.
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## Chapter 10: Further development of Complex Function Theory
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### Simple connectedness
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#### Definition (non-standard) simply connected
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Let $\Omega$ be a domain in $\mathbb{C}$. We say $\Omega$ is simply connected if $\overline{\mathbb{C}}\setminus \Omega$ is connected. ($\overline{\mathbb{C}}=\mathbb{C}\cup \{\infty\})$
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Example:
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disk is simply connected.
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annulus is not simply connected.
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$\mathbb{C}$ is simply connected.
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Any convex domain is simply connected.
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> Standard definition: $\Omega$ is simply connected if every closed curve in $\Omega$ is null-homotopic in $\Omega$.
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#### Theorem of equivalent definition of simply connected
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For open connected subsets of $\mathbb{C}$, the standard definition and the non-standard definition are equivalent.
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Proved end of book.
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#### Proposition for simply connected domain
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$\Omega$ is simply connected $\iff$ every contour in $\Omega$ has winding number $0$ about every point in $\mathbb{C}\setminus \Omega$.
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Proof:
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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 $\mathbb{C}\setminus \Omega$.
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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 $\infty\in L$.
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Let $H=\Omega\cup K=\mathbb{C}\setminus L$.
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$H$ is open, $K$ is compact subset of $H$, so by [Separation Lemma](https://notenextra.trance-0.com/Math416/Math416_L23#separation-lemma), $\exists \gamma\in H\setminus K=\Omega$ such that $K\subset \operatorname{int}(\gamma)$.
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#### Theorem 10.3 Cauchy's Theorem for simply connected domain
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corollary of Proposition for simply connected domain
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Let $\Omega$ be a simply connected domain, let $\gamma$ be a closed curve in $\Omega$. Then $\int_{\gamma} f(z) \, dz = 0$ for all $f\in O(\Omega)$.
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Proof:
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Know that is true if $\operatorname{ind}_{\gamma}(z)=0$ for all $z\in \mathbb{C}\setminus \Omega$.
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By Proposition, $\Omega$ is simply connected $\iff$ every closed curve in $\Omega$ has winding number $0$ about every point in $\mathbb{C}\setminus \Omega$.
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So the result is true.
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QED
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#### Theorem 10.4-6
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The following condition are equivalent:
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1. $\Omega$ is simply connected.
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2. every holomorphic function on $\Omega$ has a primitive $g$, i.e. $g'(z)=f(z)$ for all $z\in \Omega$.
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3. every non-vanishing holomorphic function on $\Omega$ has a holomorphic logarithm.
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4. every harmonic function on $\Omega$ has a harmonic conjugate.
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Proof:
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$(1)\iff (2)$:
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First we show $(1)\implies (2)$.
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Assume $\Omega$ is simply connected.
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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.
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$\frac{g(z+h)-g(z)}{h}=\frac{1}{h}\left[\int_{z}^{z+h} f(w) \, dw\right]$
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$\frac{1}{h}\left[\int_{z}^{z+h} f(w) \, dw\right]\to f(z)$ as $h\to 0$.
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So on $[z,z+h]\subset \Omega$, if $|f(w)-f(z)|<\epsilon$, then $|\frac{g(z+h)-g(z)}{h}-hf(z)|<h\epsilon$.
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To show $(2)\implies (1)$, we prove $\neg (1)\implies \neg (2)$.
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$(1)\iff (3)$:
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If $\Omega$ is not simply connected, there is some closed curve $\gamma$ and some $z_0\not in \Omega$ such that $\operatorname{ind}_{\gamma}(z_0)\neq 0$.
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SO $\int_{\gamma} \frac{1}{z-z_0} \, dz\neq 0$.
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So $\frac{1}{z-z_0}$ does not have a primitive on $\Omega$. $\frac{1}{z-z_0}$ have no logarithm on $\Omega$.
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This shows $(3)\implies (1)$.
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Suppose $\Omega$ is simply connected.$f\in O(\Omega)$ and $f$ is non-vanishing. We want to show that $f$ has a logarithm on $\Omega$.
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Let $z_0\in \Omega$ be fixed. And $a\in \log f(z_0)$.
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$\frac{f'(z)}{f(z)}=g'$ Since $g\in O(\Omega)$, $we can assume $g(z_0)=a$.
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$g(z)=a+\int_{z_0}^{z} g'(w) \, dw$
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So
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$$
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(fe^{-g})'=f'e^g+fe^g g'=f(e^g)'=0
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$$
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So $fe^{-g}=c$ for some $c\in \mathbb{C}$.
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So $f=ce^g$
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QED
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Continue on Residue Theorem on Thursday.
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