121 lines
3.6 KiB
Markdown
121 lines
3.6 KiB
Markdown
# Math416 Lecture 22
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## Chapter 9: Generalized Cauchy Theorem
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### Winding numbers
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Definition:
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Let $\gamma:[a,b]\to\mathbb{C}$ be a closed curve. The **winding number** of $\gamma$ around $z\in\mathbb{C}$ is defined as
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$$
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\frac{1}{2\pi i}\Delta(arg(z-z_0),\gamma)
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$$
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where $\Delta(arg(z-z_0),\gamma)$ is the change in the argument of $z-z_0$ along $\gamma$.
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#### Interior of curve
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The interior of $\gamma$ is the set of points $z\in\mathbb{C}$ such that the winding number of $\gamma$ around $z$ is non-zero.
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$$
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int_\gamma(z)=\{z\in\mathbb{C}|\frac{1}{2\pi i}\Delta(arg(z-z_0),\gamma)\neq 0\}
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$$
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#### Contour
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The winding number of a contour $\Gamma$ around $z$ is the sum of the winding numbers of the contours $\gamma_j$ around $z$.
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$$
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ind_\Gamma(z)=\sum_{j=1}^nn_j ind_{\gamma_j}(z)
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$$
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A contour is simple if $ind_\gamma(z)=\{0,1\}$ for all $z\in\mathbb{C}\setminus\gamma([a,b])$.
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#### Separation lemma
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Let $\Omega\subseteq \mathbb{C}$ be open, let $K\subset \Omega$ be compact, then $\exists$ a simple contour $\Gamma\subset \Omega\setminus K$ such that
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$$
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K\subset int_\Gamma(\Gamma)\subset \Omega
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$$
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Proof:
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First we show that $\exists$ a simple contour $\Gamma\subset \Omega\setminus K$
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Let $0<\delta<dist(K,\partial\Omega)$.
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We draw a grid fo horizontal ad vertical lines each separated from each other by $\delta$.
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Let $S_1,S_2,\dots,S_n$ be the squares that intersect $K$.
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Let $\sigma_j$ be the boundary of $S_j$ traversed in counterclockwise direction.
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Let $\varepsilon$ be the set of edges with exactly one $s_j$ for $j=1,2,\dots,q$.
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Note that $\varepsilon\subseteq \Omega\setminus K$.
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We claim that $\varepsilon$ forms a contour.
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Proof of Claim:
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Say a sequence of edges $E_1,E_2,\dots,E_p$, $E_i\in \varepsilon$. from a chain if terminal points of $E_k$ is the initial point of $E_{k+1}$ for $1\leq k\leq p-1$.
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Say it forms a cycle if inaddition the terminal points of $E_p$ is the initial point of $E_1$.
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Any cycle is a piecewise continuous closed curve.
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We want to show that $\varepsilon$ is a disjoint union of cycles.
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We can prove that every terminal point of an edge in $\varepsilon$ is an initial point of an edge in $\varepsilon$. By case analysis for the state of the four square around the terminal point.
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Let $\gamma=(E_1,E_2,\dots,E_p)$ be a maximal cycle in $\varepsilon$. (Maximal means that we cannot add another edge to it while still having a cycle.)
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Then $\gamma$ is a cycle.
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Look at the terminal point of $E_p$, This is initial point for some edge $E'$, where $E'$ is one of the edges of $\gamma$.
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If $E'$ is not $E_1$, then we can add $E'$ to $\gamma$ to form a larger cycle. Contradiction. (You can do this by case analysis. If there is three edges, then there must be four.)
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Thus $E'=E_1$.
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Thus $\gamma$ is a cycle.
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We can now remove $\gamma$ from $\varepsilon$ to form a new set $\varepsilon'$.
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We can repeat this process to form a disjoint union of cycles using induction.
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Second, we show that $int_\Gamma(\Gamma)\subset \Omega$.
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Let $z_0\in int(S_j)$ for some $1\leq j\leq q$.
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$ind_{S_k}(z_0)=\begin{cases}
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1 & k=j\\
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0 & k\neq j
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\end{cases}$
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Thus
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$$
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\begin{aligned}
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\sum_{k=1}^q ind_{S_k}(z_0)&=\sum_{k=1}^q \frac{1}{2\pi i}\int_{\partial S_k}\frac{1}{z-z_0}dz\\
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&=1\\
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&=ind_\Gamma(z_0)
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\end{aligned}
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$$
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So if $z_0\in int(\bigcup_{j=1}^q S_j)$, then $ind_\Gamma(z_0)=1$.
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And $\bigcup_{j=1}^q S_j\supset K$, so $z_0\in int_\Gamma(K)$.
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Let $z_1\in\mathbb{C}\setminus\left(\bigcup_{j=1}^q S_j\cup\Gamma\right)$.
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Then $ind_{S_k}(z_1)=0$ for all $1\leq k\leq q$.
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Thus $\sum_{k=1}^q ind_{S_k}(z_1)=0=ind_\Gamma(z_1)$.
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QED
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Continue on Generalized Cauchy Theorem next time!!
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