90 lines
2.2 KiB
Markdown
90 lines
2.2 KiB
Markdown
# Math416 Lecture 23
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## Chapter 9: Generalized Cauchy Theorem
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### Separation lemma
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Let $\Omega$ be an open subset in $\mathbb{C}$, let $K\subset \Omega$ be compact. Then There exists a simple contour $\Gamma$ such that
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$$
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K\subset \text{int}(\Gamma)\subset \Omega
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$$
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#### Corollary 9.9 for separation lemma
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Let $\Gamma$ be the contour constructed in the separation lemma. Let $f\in O(\Omega)$ be holomorphic on $\Omega$. Then $\forall z_0\in K$ such that
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$$
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f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}dz
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$$
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Proof:
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Suppose $h\in O(G)$, then $\int_{\partial S} h(z)dz=0$, by Cauchy's theorem for square, followed from the triangle case.
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So $\int_{\Gamma} h(z)dz=0=\sum_{j=1}^n \int_{\partial S_j} h(z)dz$
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Fix $z_0\in K$,
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$$
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g(z_0)=\begin{cases}
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\frac{f(z)-f(z_0)}{z-z_0} & z\neq z_0 \\
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f'(z_0) & z=z_0
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\end{cases}
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$$
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So $\int_{\Gamma} g(z)dz=0$
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Thus
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$$
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\begin{aligned}
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\int_{\Gamma}\frac{f(z)}{z-z_0}dz-\int_{\Gamma}\frac{f(z_0)}{z-z_0}dz&=0 \\
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\int_{\Gamma}\frac{f(z)}{z-z_0}dz&=f(z_0)\int_{\Gamma}\frac{1}{z-z_0}dz \\
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&=f(z_0)\cdot 2\pi i
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\end{aligned}
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$$
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QED
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#### Theorem 9.10 Cauchy's Theorem
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Let $\Omega$ be an open subset in $\mathbb{C}$, let $\Gamma$ be a contour with $int(\Gamma)\subset \Omega$. Let $f\in O(\Omega)$ be holomorphic on $\Omega$. Then
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$$
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\int_{\Gamma} f(z)dz=0
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$$
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Proof:
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Let $K\subset \mathbb{C}\setminus \text{ext}(\Gamma)$.
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By separation lemma, $\exists \Gamma_1$ s.t. $K\subset \text{int}(\Gamma_1)\subset \Omega$.
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Notice that Separation lemma ensured that $w\neq z$ for all $w\in \Gamma_1, z\in \Gamma$.
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By Corollary 9.9, $\forall z\in K, f(z)=\frac{1}{2\pi i}\int_{\Gamma_1}\frac{f(w)}{w-z}dw$
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$$
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\int_{\Gamma} f(z)dz=\frac{1}{2\pi i}\int_{\Gamma}\left[\int_{\Gamma_1}\frac{f(w)}{w-z}dw\right]dz
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$$
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By Fubini's theorem (In graduate course for analysis),
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$$
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\begin{aligned}
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\int_{\Gamma} f(z)dz&=\frac{1}{2\pi i}\int_{\Gamma_1}\left[\int_{\Gamma}\frac{f(w)}{w-z}dz\right]dw \\
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&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\left[\int_{\Gamma}\frac{1}{w-z}dz\right]dw \\
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&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\cdot 2\pi i \ \text{ind}_{\Gamma}(w)dw \\
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&=0
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\end{aligned}
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$$
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Since the winding number for $\Gamma$ on $w\in \Gamma_1$ is 0. ($w$ is outside of $\Gamma$)
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QED
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### Homotopy
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