added contour integrals
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README.md
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README.md
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@ -28,9 +28,8 @@ The topics covered in this script will be:
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5.1 Metric and Normed spaces
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5.2 Sequences, Series and Limits
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5.3 Open and Closed Sets
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5.4 ????
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5.5 Continuiuty
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5.6 Convergence of Function Sequences
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5.4 Continuiuty
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5.5 Convergence of Function Sequences
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6. Differential Calculus for Functions with multiple Variables
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6.1 Partial and Total Differentiability
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@ -42,9 +41,8 @@ The topics covered in this script will be:
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7.1 Contents and Measures
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7.2 Integrals
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7.3 Integrals over the real numbers
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7.4 ????
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7.5 Product Measures and the Fubini Theorem
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7.6 The Transformation Theorem
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7.4 Product Measures and the Fubini Theorem
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7.5 The Transformation Theorem
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8. Ordinary Differential Equations
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8.1 Solution Methods
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@ -7,5 +7,5 @@
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\pagebreak
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\subfile{sections/complex_diff.tex}
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\subfile{sections/complex_line_int.tex}
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\subfile{sections/contour_integrals.tex}
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\end{document}
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@ -1,6 +0,0 @@
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% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Complex Line Integrals}
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\end{document}
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510
chapters/sections/contour_integrals.tex
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510
chapters/sections/contour_integrals.tex
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@ -0,0 +1,510 @@
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% !TeX root = ../../script.tex
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\documentclass[../../script.tex]{subfiles}
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\begin{document}
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\section{Contour Integrals}
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\begin{defi}[Contour integrals]
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Let $U \subset \cmpln$ be open, $\gamma = C([a, b], U)$ a curve in $U$ and $f: U \rightarrow \cmpln$ continuous. Then
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\[
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\int_{\gamma} f(z) \dd{z} := \int_a^b f(\gamma(t)) \gamma'(t) \dd{t}
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\]
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\end{defi}
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\begin{eg}
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Consider the path
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\[
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\gamma(t) = re^{it}, ~\quad t \in [0, 2\pi], r > 0
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\]
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we want to take the contout integral along the path $\gamma$ of the function $z^n$
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\begin{align*}
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\int_{\gamma} z^n \dd{z} &= \int_0^{2\pi} (r e^{it})^n ire^{it} \dd{t}\\
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&= ir^{n+1} \int_0^{2\pi} e^{it(n+1)} \dd{t} = ir^{n+1} \begin{cases}
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2\pi, & n = -1 \\
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0, & n \ne -1
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\end{cases}
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\end{align*}
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\end{eg}
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\begin{lem}[Estimation Lemma]
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For every curve $\gamma \in C([0, 1], U)$ and every continuous function $f: U \rightarrow \cmpln$ we have
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\[
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\abs{\int_{\gamma} f(z) \dd{z}} \le \sup_{z \in \gamma} \abs{f(z)} \int_0^1 \abs{\gamma'(t)} \dd{t}
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\]
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\end{lem}
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\begin{proof}
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\begin{equation}
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\begin{split}
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\abs{\int_{\gamma} f(z) \dd{z}} = \abs{\int_0^1 f(\gamma(t)) \gamma'(t) \dd{t}} &\le \int_0^1 \abs{f(\gamma(t))} \abs{\gamma'(t)} \dd{t} \\
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&\le \sup_{t \in [0, 1]} \abs{f(\gamma(t))} \int_0^1 \abs{\gamma'(t)} \dd{t}
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\end{split}
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\end{equation}
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\end{proof}
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\begin{cor}
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Let $\gamma \in C([0, 1], U)$ be a simple closed curve, $U \subset \cmpln$, and let $f: U \rightarrow \cmpln$ a holomorphic function with
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\begin{align*}
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u = \Re f && v = \Im f
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\end{align*}
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Then
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\[
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\oint_{\gamma} f(z) \dd{z} = 0
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\]
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\end{cor}
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\begin{proof}
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Let $A \subset U$ be the surface bounded by $\gamma$. Then
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\begin{equation}
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\oint_{\gamma} f(z) \dd{z} = \int_0^1 f(\gamma(t)) \gamma'(t) \dd{t}
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\end{equation}
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We can split $\gamma$ into a real and an imaginary part, like this
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\begin{equation}
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\gamma(t) = \gamma_1(t) + i\gamma_2(t), \quad \gamma_1, \gamma_2: [0, 1] \rightarrow \realn
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\end{equation}
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Then we can calculate
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\begin{equation}
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\begin{split}
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\oint_{\gamma} f(z) \dd{z} &= \int_0^1 \left(u(\gamma_1(t), \gamma_2(t)) + iv(\gamma_1(t), \gamma_2(t)) (\gamma_1'(t) + i\gamma_2'(t))\right) \dd{t} \\
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&= \int_0^1 u(\gamma_1(t), \gamma_2(t)) \gamma_1'(t) - v(\gamma_1(t), \gamma_2(t)) \gamma_2'(t) \dd{t} \\
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&\quad + i\int_0^1 u(\gamma_1(t), \gamma_2(t))\gamma_2'(t) + v(\gamma_1(t), \gamma_2(t))\gamma_1'(t) \dd{t} \\
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&= \int_0^1 \begin{pmatrix}
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u(\gamma(t)) \\ -v(\gamma(t))
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\end{pmatrix}
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\begin{pmatrix}
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\gamma_1'(t) \\ \gamma_2'(t)
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\end{pmatrix}
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\dd{t} + i \int_0^1 \begin{pmatrix}
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v(\gamma(t)) \\ u(\gamma(t))
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\end{pmatrix}
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\begin{pmatrix}
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\gamma_1'(t) \\ \gamma_2'(t)
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\end{pmatrix}
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\dd{t} \\
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&= \oint_{\boundary{A}} \begin{pmatrix}
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u \\ -v
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\end{pmatrix} \dd{s} + i \oint_{\boundary{A}} \begin{pmatrix}
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v \\ u
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\end{pmatrix} \\
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&= \int_A (-\partial_x v - \partial_y u) \dd{\lambda^2} + i \int_A (\partial_x u - \partial_y v) \dd{\lambda^2} \\
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\end{split}
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\end{equation}
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Because $f$ is holomorphic we can apply the Cauchy-Riemann equation
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\begin{equation}
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\oint_{\gamma} f(z) \dd{z} = 0
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\end{equation}
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\end{proof}
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\begin{defi}
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\begin{enumerate}[(i)]
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\item A closed curve $\gamma: [a, b] \rightarrow U$ with $U \subset \cmpln$ is said to be null-homotopic,
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if it can be continuously deformed into a point within the set $U$.
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\item Two curves $\gamma_1, \gamma_2: [0, 1] \rightarrow U$ with identical boundary points
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\begin{align*}
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\gamma_1(0) = \gamma_2(0) ~\wedge~ \gamma_1(1) = \gamma_2(1)
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\end{align*}
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is said to be homotopic in $U$ if the concatenation
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\begin{align*}
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\gamma: [0, 2] &\longrightarrow U \\
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\gamma(t) &= \begin{cases}
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\gamma_1(t), & t \in [0, 1] \\
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\gamma_2(2 - t) & t \in [1, 2]
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\end{cases}
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\end{align*}
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is null-homotopic.
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\item Two closed surves $\gamma_0, \gamma_1$ are said to be free-homotopic in $U$ if they can be continuously transformed into each other.
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\end{enumerate}
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\end{defi}
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\begin{defi}
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A non-empty set $U \subset \cmpln$ is said to be
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\begin{enumerate}[(i)]
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\item \textit{connected} if any two points in $U$ can be connected by a curve in $U$.
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\item \textit{simply connected} if $U$ is connected and every closed surve in $U$ is null-homotopic.
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\item a \textit{domain} if it is open and connected.
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\end{enumerate}
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\end{defi}
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\begin{thm}[Cauchy's Integral Theorem]
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Let $f: U \rightarrow \cmpln$ be holomorphic and $\gamma$ a closed, null-homotopic curve in $U \subset \cmpln$ open. Then
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\[
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\oint_{\gamma} f(z) \dd{z} = 0
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\]
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\end{thm}
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\begin{proof}
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\noproof
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\end{proof}
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\begin{cor}
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\begin{enumerate}[(i)]
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\item Let $\gamma_1, \gamma_2$ be holomorphic curves with the same endpoints on the open set $U \subset \cmpln$. Then
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\[
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\int_{\gamma_1} f(z) \dd{z} = \int_{\gamma_2} f(z) \dd{z}
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\]
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for all holomorphic $f: U \rightarrow \cmpln$.
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\item For $f: U \rightarrow \cmpln$ holomorphic, with $U \subset \cmpln$ open and simple connected. Then $\forall z_0 \in U$
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\[
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F(z) := \int_{z_0}^{z} f(\zeta) \dd{\zeta} = \int_{\gamma = \gamma_0} f(\zeta) \dd{\zeta}
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\]
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is a holomorphic anti-derivative of $f$, i.e.
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\[
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F'(z) = f(z) \quad \forall z \in U
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\]
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\end{enumerate}
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\end{cor}
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\begin{proof}
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First we prove (i). The concatenation $\gamma := \gamma_1 \gamma_2$ is a null-homotopic curve, so together with the holomorphy of $f$ we can apply the Cauchy integral theorem
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\begin{equation}
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\begin{split}
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0 = \oint_{\gamma} f(z) \dd{z} &= \int_0^2 f(\gamma(t)) \dot{\gamma}(t) \dd{t} \\
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&= \int_0^1 f(\gamma_1(t))\dot{\gamma_1}(t) \dd{t} - \int_1^2 f(\gamma_2(2 - t)) \dot{\gamma_2}(2 - t) \dd{t}
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\end{split}
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\end{equation}
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Substitute $s = 2 - t$ with $\dd{s} = -\dd{t}$:
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\begin{equation}
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\begin{split}
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&= \int_0^1 f(\gamma_1(t)) \dot{\gamma}(t) \dd{t} - \gamma_0^1 f(\gamma_2(s)) \dot{\gamma_2}(s) \dd{s} \\
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&= \int_{\gamma_1} f(z) \dd{z} - \int_{\gamma_2} f(z) \dd{z}
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\end{split}
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\end{equation}
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Now we prove (ii). According to (i), we have
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\begin{equation}
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F(z + h) = F(z) + \int_{\gamma_{z+h, z}} f(z) \dd{z}
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\end{equation}
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We choose $\gamma_{z+h, z}$ to be a straight line, i.e.
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\begin{equation}
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\gamma_{z+h, z}(t) = t(z + h) + (1 - t)z, \quad t \in [0, 1]
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\end{equation}
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Then
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\begin{equation}
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\int_{\gamma_{z+h, z}} 1 \dd{\zeta} = \int_0^1 \dot{\gamma}(t) \dd{t} = h
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\end{equation}
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Thus follows
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\begin{equation}
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F(z + h) - F(z) = \int_{\gamma_{z + h, z}} f(\zeta) \dd{\zeta} \iff \frac{F(z+h) - F(z)}{h} = \frac{1}{h} \in f(\zeta) \dd{\zeta}
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\end{equation}
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and therefore
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\begin{equation}
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\begin{split}
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\left| \frac{F(z+h) - F(z)}{h} - f(z) \right| = &\left| \frac{1}{h} \int_{\gamma_{z+h, z}} f(\zeta) \dd{\zeta} - f(z) \right| \\
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= &\left| \frac{1}{h} \int_{\gamma_{z+h, z}} f(\zeta) - f(z) \dd{\zeta} \right| \\
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= &\frac{1}{\abs{h}} \int_0^1 \left| f(\gamma_{z+h, z}(t)) - f(z)\right| \left| \dot{\gamma_{z+h, z}}(t) \right| \dd{t} \\
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\le &\frac{1}{h} \sup_{t \in [0, 1]} \left| f(\gamma_{z+h,z}(t)) - f(z) \right| \cdot \underbrace{\int \left| \dot{\gamma_{z+h, z}}(t) \right| \dd{t}}_{\abs{h}} \\
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= &\sup_{t \in [0, 1]} \left| f(\gamma_{z+h, z}(t)) - f(z) \right| \\
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\conv{k \rightarrow 0} &0
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\end{split}
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\end{equation}
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\end{proof}
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\begin{eg}[The complex logarithm]
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Consider $t \mapsto e^{it}, ~t \in \realn$. This is a $2\pi$-periodic function, that means
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\[
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e^{it} = e^{i(t + 2\pi n)}, \quad n \in \intn
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\]
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The function
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\begin{align*}
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f: \cmpln \setminus \set{0} &\longrightarrow \cmpln \\
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z &\longmapsto \frac{1}{z}
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\end{align*}
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is holomorphic, and does not have an anti-derivative on $\cmpln \setminus \set{0}$. If it did, then
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\[
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\int_{\gamma} f(z) \dd{z} = F(\gamma(2\pi)) - F(\gamma(0)) = 0
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\]
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would have to hold, but we know that
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\[
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\int_{\gamma} \frac{\dd{z}}{z} = 2\pi i
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\]
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This is a contradiction. However $f$ does have an anti-derivative on $\cmpln_{-}$ (the complex numbers without the negative real axis) , since $\cmpln_{-}$ is simple connected and $f$ is holomorphic.
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Thus we can define
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\begin{align*}
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\Log: \cmpln_{-} &\longrightarrow \cmpln \\
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z &\longmapsto \int_{\gamma: [0, 1] \rightarrow z} \frac{\dd{\zeta}}{\zeta}
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\end{align*}
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It can also be defined as
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\begin{align*}
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\Log z = \begin{cases}
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0, & 1 \\
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\log\abs{z} + i \arg(z), & \text{else}
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\end{cases}
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\end{align*}
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The function $\arg$ is defined as
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\begin{align*}
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\arg: \cmpln_{-} &\longrightarrow (-\pi, \pi) \\
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z &\longmapsto \phi \text{ for } z = \abs{z} e^{i\phi}
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\end{align*}
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$\Log$ is said to be the main branch of the complex logarithm, and
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\[
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\Log z = \log\abs{z} + i (\arg(z) + 2\pi n), \quad n \in \intn
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\]
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the secondary branches.
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\end{eg}
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\begin{eg}[Fresnel Integrals]
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Consider the integrals
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\begin{align*}
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\int_0^{\infty} \cos(t^2) \dd{t} && \int_0^{\infty} \sin(t^2) \dd{t}
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\end{align*}
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The way these integrals are supposed to be interpreted is as
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\[
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\int_0^{\infty} f(t) \dd{t} = \lim_{N \rightarrow \infty} \int_0^N f(t) \dd{t}
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\]
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We realize that
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\begin{align*}
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\cos(t^2) = \Re e^{-it^2} && \sin(t^2) = -\Im e^{-it^2}
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\end{align*}
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Now, consider these paths
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\begin{center}
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\begin{tikzpicture}
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\draw[->, >=stealth] (0, -1) -- (0, 5) node[above] {$\Im$};
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\draw[->, >=stealth] (-1, 0) -- (5, 0) node[right] {$\Re$};
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\draw[->, >=stealth, very thick] (0, 0) -- node[below] {$\gamma_1$} (4, 0) node[below] {$R$};
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\draw[->, >=stealth, very thick] (4, 0) -- node[right] {$\gamma_2$} (4, 4);
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\draw[->, >=stealth, very thick] (0, 0) -- node[above left] {$\gamma$} (4, 4);
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\draw[dashed] (0, 4) node[left] {$R$} -- (4, 4);
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\end{tikzpicture}
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\end{center}
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So it becomes apparent that
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\[
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\int_0^R \cos(t^2) \dd{t} = \Re \int_0^R e^{-it^2} \dd{t} = \Re \int_{\gamma} e^{-z^2} \dd{z}
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\]
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We can define a new (closed) path
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\[
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\Gamma = \gamma_1 \gamma_2 (-\gamma)
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\]
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and with Cauchy's theorem we can realize that
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\begin{align*}
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0 = \oint_{\Gamma} e^{-z^2} \dd{z} = \int_{\gamma_1} e^{-z^2} \dd{z} + \int_{\gamma_2} e^{-z^2} \dd{z} - \int_{\gamma} e^{-z^2} \dd{z}
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\end{align*}
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The next step is to evaluate each of the integrals in the last term, starting with the integral over $\gamma$.
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\begin{align*}
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\int_{\gamma} e^{-z^2} \dd{z} = &\int_0^R e^{-((1+i)t)^2} (1+i) \dd{t} \\
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= &(1 + i) \int_0^R e^{-2it^2} \dd{t} \\
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= &\frac{1+i}{\sqrt{2}} \int_0^{\sqrt{2} R} e^{-is^2} \dd{s}
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\end{align*}
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The integrall over $\gamma_1$ evaluates to
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\begin{align*}
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\int_{\gamma_1} e^{-z^2} \dd{z} = \int_0^R e^{-t^2} \dd{t} \conv{R \rightarrow \infty} \int_0^{\infty} e^{-t^2} \dd{t} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-t^2} \dd{t} = \frac{\sqrt{\pi}}{2}
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\end{align*}
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And the one over $\gamma_2$ to
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\begin{align*}
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\int_{\gamma_2} e^{-z^2} \dd{z} = \int_0^R e^{-(r +it)^2} i \dd{t} = i \int_0^R e^{-R^2 + t^2} e^{-2irt} \dd{t}
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\end{align*}
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To evaluate this we need to consider the absolute value of this integral
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\begin{align*}
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\abs{\int_{\gamma_2} e^{-z^2} \dd{z}} \le &e^{-R^2} \int_0^R e^{t^2} \underbrace{\abs{e^{-2iRt}}}_{=1} \dd{t} \\
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= &e^{-R^2} \int_0^R e^{t^2} \dd{t} \le e^{-R^2} \int_0^R e^{tR} \dd{t} \\
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= &e^{-R^2} \left[\frac{1}{R} e^{tR}\right]_0^R = \frac{e^{-R^2}}{R} \left(e^{R^2} - 1\right)
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\end{align*}
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so
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\[
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\abs{\int_{\gamma_2} e^{-z^2} \dd{z}} \le \frac{1}{R} \left(1 - e^{-R^2}\right) \conv{R \rightarrow \infty} 0
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\]
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Thus we can calculate
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\[
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\int_{\gamma} e^{-z^2} \dd{z} = \frac{1+i}{\sqrt{2}} \int_0^{\sqrt{2} R} e^{-it^2} \dd{t} = \int_{\gamma_1} e^{-z^2} \dd{z} + \int_{\gamma_2} e^{-z^2} \dd{z}
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\]
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And finally
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\begin{align*}
|
||||
\lim_{R \rightarrow \infty} \int_0^{\infty} e^{-it^2} \dd{t} = &\lim_{R \rightarrow \infty} \int_0^{\sqrt{2} R} e^{-t^2} \dd{t} \\
|
||||
= &\frac{\sqrt{2}}{1+i} \left(\lim_{R \rightarrow \infty} \left( \int_{\gamma_1} e^{-z^2} \dd{z} + \int_{\gamma_2} e^{-z^2} \dd{z} \right) \right) \\
|
||||
= &\frac{\sqrt{2}}{1+i} \left(\frac{\pi}{2} + 0\right) \\
|
||||
= &\sqrt{\frac{\pi}{2}} \frac{1-i}{2} = \sqrt{\frac{\pi}{8}} (1 - i)
|
||||
\end{align*}
|
||||
So we can calculate the Fresnel integrals
|
||||
\begin{align*}
|
||||
\int_0^{\infty} \cos(t^2) \dd{t} &= \sqrt{\frac{\pi}{8}} \\
|
||||
\int_0^{\infty} \sin(t^2) \dd{t} &= \sqrt{\frac{\pi}{8}}
|
||||
\end{align*}
|
||||
\end{eg}
|
||||
|
||||
\begin{thm}[Cauchy's Theorem for circular disks]
|
||||
Let $f: U \rightarrow \cmpln$ be holomorphic, $U \subset \cmpln$ open and $\cball[r](a) \subset U$. Then
|
||||
\[
|
||||
f(a) = \frac{1}{2\pi i} \int_{\abs{z - a} = r} \frac{f(z)}{z-a} \dd{z}
|
||||
\]
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
Consider the following path
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=7.5]
|
||||
\draw[fill] (5, 0.7) node[below] {$a$} circle [radius=0.004];
|
||||
|
||||
\begin{scope}[decoration={
|
||||
markings,
|
||||
mark=at position 0.5 with {\arrow{latex}}}
|
||||
]
|
||||
\draw[thick, postaction={decorate}, domain=435:105] plot ({0.1*cos(\x) + 5}, {0.1*sin(\x) + 0.7});
|
||||
\draw[postaction={decorate}, domain=435:105] plot ({0.3*cos(\x) + 5}, {0.3*sin(\x) + 0.7});
|
||||
\draw[thick, postaction={decorate}] ({0.1*cos(105) + 5}, {0.1*sin(105) + 0.7}) -- node[left] {$\alpha_{\delta}^1$} ({0.3*cos(105) + 5}, {0.3*sin(105) + 0.7});
|
||||
\draw[thick, postaction={decorate}] ({0.3*cos(435) + 5}, {0.3*sin(435) + 0.7}) -- node[right] {$\alpha_{\delta}^2$} ({0.1*cos(435) + 5}, {0.1*sin(435) + 0.7});
|
||||
|
||||
\draw[thick, postaction={decorate}, domain=105:75] plot ({0.3*cos(\x) + 5}, {0.3*sin(\x) + 0.7});
|
||||
\end{scope}
|
||||
|
||||
\node[below] at ({0.1*cos(270) + 5}, {0.1*sin(270) + 0.7}) {$\gamma_{\epsilon}$};
|
||||
\node[below] at ({0.3*cos(270) + 5}, {0.3*sin(270) + 0.7}) {$|z - a| = r$};
|
||||
|
||||
\draw[dashed] (5, 0.7) -- ({0.1*cos(150) + 5}, {0.1*sin(150) + 0.7}) node[above left=-0.1cm] {$\epsilon$} ;
|
||||
\draw[dashed] (5, 0.7) -- ({0.3*cos(5) + 5}, {0.3*sin(5) + 0.7}) node[right] {$r$};
|
||||
|
||||
\node[above] at ({0.3*cos(90) + 5}, {0.3*sin(90) + 0.7}) {$\gamma_{\delta}$};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
According to the first corollary of Cauchy's theorem we have
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\int_{\abs{z-a}=r} \frac{f(z)}{z-a} \dd{z} = &\lim_{\delta \rightarrow 0} \int_{\gamma_{\epsilon, \delta}} \frac{f(z)}{z-a} \dd{z} + \underbrace{\int_{\alpha_{\delta}^1} \frac{f(z)}{z-a} \dd{z} + \int_{\alpha_{\delta}^2} \frac{f(z)}{z-a} \dd{z}}_{\conv{\delta \rightarrow 0} 0} \\
|
||||
= &\int_{\gamma_{\epsilon}} \frac{f(z)}{z-a} \dd{z}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
Thus we conclude
|
||||
\begin{equation}
|
||||
\begin{split}
|
||||
\int_{\abs{z-a}=r} \frac{f(z)}{z-a} \dd{z} = \int_{\gamma_{\epsilon}} \frac{f(z)}{z-a} \dd{z} = &\int_{\gamma_{\epsilon}} \frac{f(z) - f(a)}{z-a} \dd{z} + \int_{\gamma_{\epsilon}} \frac{f(a)}{z-a} \dd{z} \\
|
||||
= &\int_{\gamma_{\epsilon}} \frac{f(z) - f(a)}{z-a} \dd{z} + f(a) \int_{\gamma_{\epsilon}} \frac{\dd{z}}{z-a}
|
||||
\end{split}
|
||||
\end{equation}
|
||||
We also know that
|
||||
\begin{equation}
|
||||
\int_{\gamma_{\epsilon}} \frac{\dd{z}}{z-a} = 2\pi i
|
||||
\end{equation}
|
||||
Since $f$ is holomorphic we can realize
|
||||
\begin{equation}
|
||||
\sup_{\cball[r](a)} \left| \frac{f(z) - f(a)}{z - a} \right| = M_r < \infty
|
||||
\end{equation}
|
||||
Which results in
|
||||
\begin{equation}
|
||||
\abs{\int_{\gamma_{\epsilon}} \frac{f(z) - f(a)}{z - a} \dd{z}} \le M_r \underbrace{\int_0^{2\pi} \abs{\dot{\gamma_{\epsilon}}(t)} \dd{t}}_{2\pi\epsilon} \conv{\epsilon \rightarrow 0} 0
|
||||
\end{equation}
|
||||
Thus follows
|
||||
\begin{equation}
|
||||
\int_{|z - a| = r} \frac{f(z)}{z - a} \dd{z} = \int_{\gamma_{\epsilon}} \frac{f(z) - f(a)}{z - a} \dd{z} + 2\pi i f(a) \conv{\epsilon \rightarrow 0} 2\pi i f(a)
|
||||
\end{equation}
|
||||
Or short
|
||||
\begin{equation}
|
||||
\int_{\abs{z - a} = r} \frac{f(z)}{z - a} \dd{z} = 2\pi i f(a)
|
||||
\end{equation}
|
||||
\end{proof}
|
||||
|
||||
\begin{cor}
|
||||
Let $f: U \rightarrow \cmpln$ be a holomorphic function and $U \subset \cmpln$ an open set such that $\cball[r](a) \subset U$. Then
|
||||
\[
|
||||
f(a) = \frac{1}{2\pi} \int_0^{2\pi} f(a + re^{it}) \dd{t}
|
||||
\]
|
||||
\end{cor}
|
||||
\begin{proof}
|
||||
\reader
|
||||
\end{proof}
|
||||
|
||||
\begin{defi}[Analytic functions]
|
||||
Let $f: U \rightarrow \cmpln$ be a function and $U \subset \cmpln$ a domain. $f$ is said to be analytic in $z_0 \in U$ if and only if there exists a power series
|
||||
\[
|
||||
\sum_{n=0}^{\infty} a_n \zeta^n
|
||||
\]
|
||||
with convergence radius
|
||||
\[
|
||||
\rho = \left(\limsup |a_n|^{\frac{1}{n}}\right)^{-1} > 0
|
||||
\]
|
||||
and $\delta \in (0, \rho)$ such that $\oball[\delta](z_0) \subset U$ and
|
||||
\[
|
||||
f(z) = \int_{k=0}^{\infty} a_n (z - z_0)^n, \quad \forall z \in \oball[\delta](z_0)
|
||||
\]
|
||||
$f$ is said to be analytic on $U$ if $f$ is analytic $\forall z_0 \in U$.
|
||||
\end{defi}
|
||||
|
||||
\begin{thm}[Power series expansion]
|
||||
If $f$ is holomorphic on a circular disk $\oball[r](z_0)$ for some $r > 0$, then $f$ is analytic in $z_0$.
|
||||
$f$ can be represented with the on $\oball[\rho](z_0)$ convergent power series
|
||||
\[
|
||||
f(z) = \sum_{n=0}^{\infty} c_n (z - z_0)^n, \quad z \in \oball[\rho](z_0)
|
||||
\]
|
||||
with
|
||||
\[
|
||||
c_n = \frac{1}{2\pi i} \int_{|z - z_0| = r} \frac{f(z)}{(z - z_0)^{n+1}} \dd{z}, \quad \forall \rho \in (0, r)
|
||||
\]
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
\noproof
|
||||
\end{proof}
|
||||
|
||||
\begin{rem}
|
||||
If $f$ is holomorphic then $f$ can be infinitely often differentiated on $\cmpln$ with
|
||||
\[
|
||||
f^{(n)}(z) = n! c_n = \frac{n!}{2\pi i} \int_{\abs{z - z_0} = \rho} \frac{f(z)}{(z- z_0)^{n+1}} \dd{z}
|
||||
\]
|
||||
By employing the estimation lemma we can then find that
|
||||
\begin{align*}
|
||||
\abs{c_n} \le \frac{1}{2\pi} \abs{\int_{\abs{z - z_0} = \rho} \frac{f(z)}{(z - z_0)^{n+1}} \dd{z}} \le &\frac{1}{2\pi} \sup_{\abs{z - z_0} = \rho} \abs{\frac{f(z)}{\abs{z - z_0}^{n+1}}} \cdot 2\pi \rho \\
|
||||
= &\frac{1}{\rho^n} \sup_{z \in \oball[r](z_0)} \abs{f(z)} \\
|
||||
= &\frac{M_r}{\rho^n}, \quad M_r < \infty
|
||||
\end{align*}
|
||||
This is Cauchy's estimate.
|
||||
\end{rem}
|
||||
|
||||
\begin{thm}[Liouville's Theorem]
|
||||
Every bounded entire function is constant.
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
According to the power series expansion theorem, $f$ can be represented by a power series on all of $\cmpln$:
|
||||
\begin{equation}
|
||||
f(z) = \sum_{n=0}^{\infty} c_n z^n
|
||||
\end{equation}
|
||||
and the coefficients satisfy the Cauchy estimate
|
||||
\begin{equation}
|
||||
\abs{c_n} \le \frac{1}{\rho^n} \sup_{\abs{z} = \rho} \abs{f(z)} \le \frac{1}{\rho^n} \underbrace{\sup_{z \in \cmpln} \abs{f(z)}}_{< \infty}
|
||||
\end{equation}
|
||||
This inequality tends to $0$ if $\rho$ tends to $\infty$ for all $n \ge 1$, thus we can find
|
||||
\begin{equation}
|
||||
c_n = 0, \quad \forall n \ge 1
|
||||
\end{equation}
|
||||
Thus
|
||||
\begin{equation}
|
||||
f(z) = c_0 = \const.
|
||||
\end{equation}
|
||||
\end{proof}
|
||||
|
||||
\begin{thm}[Fundamental Theorem of Algebra]
|
||||
Every polynomial of degree $n \ge 1$
|
||||
\[
|
||||
f(z) = \sum_{k=0}^n c_k z^k, \quad c_n \ne 0
|
||||
\]
|
||||
has a root, i.e.
|
||||
\[
|
||||
\exists z_0 \in \cmpln: \quad f(z_0) = 0
|
||||
\]
|
||||
\end{thm}
|
||||
\begin{proof}
|
||||
Assume there exists no root. Then the function
|
||||
\begin{equation}
|
||||
z \longmapsto \frac{1}{f(z)}
|
||||
\end{equation}
|
||||
would be holomorphic on all of $\cmpln$, since $z \mapsto \rec{z}$ is holomorphic on $\cmpln \setminus \set{0}$.
|
||||
Furthermore we find that
|
||||
\begin{equation}
|
||||
\exists R \ge 0: ~~\abs{z} \ge R \implies \abs{f(z)} \ge \abs{f(0)} > 0
|
||||
\end{equation}
|
||||
which implies
|
||||
\begin{equation}
|
||||
\sup_{z \in \cmpln} \frac{1}{\abs{f(z)}} = \sup_{\abs{z} < R} \frac{1}{\abs{f(z)}} = \max_{\abs{z} \le R} \frac{1}{\abs{f(z)}} < \infty
|
||||
\end{equation}
|
||||
since $f$ doesn't have a root. According to Liouville's theorem $\rec{f}$ has to be constant, and thus $f$ must be constant.
|
||||
This implies that $c_n = 0$, which contradicts the assumption. So $f$ has to have a root.
|
||||
\end{proof}
|
||||
|
||||
\begin{cor}[Polynomial Decomposition]
|
||||
Let
|
||||
\[
|
||||
f(z) = \sum_{k=0}^n c_k z^k, \quad n \in \natn, c_k \in \cmpln, c_n = 1
|
||||
\]
|
||||
Then $\exists z_j \in \cmpln, ~~j = 1, \cdots, n$ such that
|
||||
\[
|
||||
f(z) = \prod_{j=1}^n (z - z_j)
|
||||
\]
|
||||
\end{cor}
|
||||
\end{document}
|
||||
11350
script.pdf
11350
script.pdf
File diff suppressed because it is too large
Load diff
|
|
@ -11,7 +11,7 @@
|
|||
\usepackage{fancyhdr}
|
||||
\usepackage{pdfpages}
|
||||
\usepackage[arrowdel]{physics}
|
||||
\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,angles,patterns}
|
||||
\usetikzlibrary{calc,trees,decorations.markings,positioning,arrows,fit,shapes,angles,patterns}
|
||||
\DeclareDocumentCommand\vnabla{}{\vectorarrow{\nabla}}
|
||||
|
||||
\graphicspath{assets}
|
||||
|
|
@ -38,6 +38,7 @@
|
|||
\DeclareMathOperator{\sgn}{sgn}
|
||||
\DeclareMathOperator{\diag}{diag}
|
||||
\DeclareMathOperator{\const}{const}
|
||||
\DeclareMathOperator{\Log}{Log}
|
||||
|
||||
\newcommand{\natn}{\mathbb{N}}
|
||||
\newcommand{\intn}{\mathbb{Z}}
|
||||
|
|
@ -86,7 +87,7 @@
|
|||
|
||||
\newcommand{\interior}[1]{\mathring{#1}}
|
||||
\newcommand{\boundary}[1]{\partial #1}
|
||||
\newcommand{\closure}[1]{\bar{#1}}
|
||||
\newcommand{\closure}[1]{\overline{#1}}
|
||||
|
||||
\newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}}
|
||||
\newcommand{\convinf}{\conv{n \rightarrow \infty}}
|
||||
|
|
|
|||
Loading…
Reference in a new issue