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# Lecture 19: Nov. 15, 2023 ## Proof of Cauchy's Integral Formula ### Cauchy's Integral Formula Let $f$ be analytic on $B(z_0;R)$ and $|z-z_0| < r < R$. Then $$ f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w-z} dw $$ where $\gamma(t) = z_0 + re^{it}$, $0 \leq t \leq 2\pi$. ### Proof Fix $z...

# Lecture 19: Nov. 15, 2023 ## Proof of Cauchy's Integral Formula ### Cauchy's Integral Formula Let $f$ be analytic on $B(z_0;R)$ and $|z-z_0| < r < R$. Then $$ f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w-z} dw $$ where $\gamma(t) = z_0 + re^{it}$, $0 \leq t \leq 2\pi$. ### Proof Fix $z \in B(z_0;r)$. Since $\frac{f(w)}{w-z}$ is analytic on $B(z_0;R) \setminus \{z\}$, we have that ($ \epsilon $ is small enough) $$ \int_{\gamma} \frac{f(w)}{w-z} dw = \int_{C_\epsilon} \frac{f(w)}{w-z} dw $$ where $C_\epsilon(t) = z + \epsilon e^{it}, 0 \leq t \leq 2\pi$. Now, $$ \int_{C_\epsilon} \frac{f(w)}{w-z} dw = \int_{C_\epsilon} \frac{f(z)}{w-z} dw + \int_{C_\epsilon} \frac{f(w)-f(z)}{w-z} dw $$ Note that $w = z + \epsilon e^{it}$, so $\int_{C_\epsilon} \frac{f(z)}{w-z} dw = f(z) \int_0^{2\pi} \frac{1}{\epsilon e^{it}} i\epsilon e^{it} dt = 2\pi i f(z)$. We want to show that $\left| \int_{C_\epsilon} \frac{f(w)-f(z)}{w-z} dw \right| \rightarrow 0$ as $\epsilon \rightarrow 0$. Since $f$ is analytic, it is continuous. So given any $\delta > 0$, there exists an $\epsilon > 0$ such that $|w-z| < \epsilon$ implies that $|f(w)-f(z)| < \delta $. Thus, $\left| \int_{C_\epsilon} \frac{f(w)-f(z)}{w-z} dw \right| \leq \int_{C_\epsilon} \frac{|f(w)-f(z)|}{|w-z|} |dw| \leq \frac{\delta}{\epsilon} 2\pi \epsilon = 2\pi \delta$. Since we can make this arbitrarily small, $\left| \int_{C_\epsilon} \frac{f(w)-f(z)}{w-z} dw \right| \rightarrow 0$ as $\epsilon \rightarrow 0$. Therefore, $$ \int_{\gamma} \frac{f(w)}{w-z} dw = 2\pi i f(z). $$ ## Consequences of Cauchy's Integral Formula Suppose $f: G \rightarrow \mathbb{C}$ is analytic. ### Theorem $f$ is infinitely differentiable on $G$. ### Proof Let $z \in G$. Then there exists a $B(z;R) \subseteq G$ such that $f$ is analytic on $B(z;R)$. Let $|w-z| < r < R$. Then $f(w) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - w} d\xi$, where $\gamma(t) = z + re^{it}, 0 \leq t \leq 2\pi$. Differentiating under the integral sign, we get $$ f'(w) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{(\xi - w)^2} d\xi $$ and $$ f''(w) = \frac{2}{2\pi i} \int_\gamma \frac{f(\xi)}{(\xi - w)^3} d\xi $$ In general, $$ f^{(n)}(w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(\xi)}{(\xi - w)^{n+1}} d\xi $$