121 lines
3.6 KiB
Markdown
121 lines
3.6 KiB
Markdown
# Math 416 Lecture 12
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## Continue on last class
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### Cauchy's Theorem on triangles
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Let $T$ be a triangle in $\mathbb{C}$ and $f$ be holomorphic on $T$. Then
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$$
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\int_T f(\zeta) d\zeta = 0
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$$
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### Cauchy's Theorem for Convex Sets
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Let's start with a simple case: $f(\zeta)=1$.
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For any closed curve $\gamma$ in $U$, we have
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$$
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\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
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$$
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#### Definition of a convex set
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A set $U$ is convex if for any two points $\zeta_1, \zeta_2 \in U$, the line segment $[\zeta_1, \zeta_2] \subset U$.
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Let $O(U)$ be the set of all holomorphic functions on $U$.
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#### Definition of primitive
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Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(\zeta)=f(\zeta)$ for all $\zeta \in U$, then $g$ is called a primitive of $f$ on $U$.
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#### Cauchy's Theorem for a Convex region
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Cauchy's Theorem holds if $f$ has a primitive on a convex region $U$.
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$$
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\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)
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$$
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Since the curve is closed, $\zeta_1=\zeta_2$, so $\int_\gamma f(\zeta) d\zeta = 0$.
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Proof:
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It is sufficient to prove that if $U$ is convex, $f$ is holomorphic on $U$, then $f=g'$ for some $g$ holomorphic on $U$.
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We pick a point $z_0\in U$ and define $g(\zeta)=\int_{[\zeta_0,\zeta]}f(\xi)d\xi$.
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We claim $g\in O(U)$ and $g'=f$.
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Let $\zeta_1$ close to $\zeta$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $\zeta\in T$ and $T\subset U$.
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$$
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\begin{aligned}
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0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
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&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
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\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
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\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
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&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
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&=I
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\end{aligned}
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$$
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Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1)$.
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There exists a $\delta>0$ such that $|\zeta-\zeta_1|<\delta$ implies $|f(\zeta)-f(\zeta_1)|<\epsilon$.
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So
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$$
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|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon
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$$
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So $I\to 0$ as $\zeta_1\to\zeta$.
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Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.
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EOP
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### Cauchy's Theorem for a disk
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Let $U$ be the open set, $f\in O(U)$. Let $C$ be a circle inside $U$ and $\zeta$ be a point inside $C$.
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Then
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$$
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f(\zeta)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-\zeta} d\xi
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$$
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Proof:
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Let $C_\epsilon$ be a circle with center $\zeta$ and radius $\epsilon$ inside $C$.
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Claim:
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$$
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\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_{C}\frac{f(\xi)d\xi}{\xi-\zeta}
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$$
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We divide the integral into four parts:
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Notice that $\frac{f(\xi)}{\xi-\zeta}$ is holomorphic whenever $f(\xi)\in U$ and $\xi\in \mathbb{C}\setminus\{\zeta\}$.
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So we can apply Cauchy's theorem to the integral on the inside square.
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$$
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\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=0
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$$
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Since $\frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-\zeta}d\xi=1$, $\sigma=\epsilon e^{it}+\zeta_0$ and $\sigma'=\epsilon e^{it}$, we have
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/* TRACK LOST*/
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$$
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\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-\zeta}=2\pi i f(\zeta)
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$$
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EOP
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