From 777f9d3fd8caf56e6bc6999a4b05379307d0733f Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Tue, 12 Sep 2023 07:36:33 +0200
Subject: Initial commit

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+---
+title: Cauchy's Integral Formula
+parent: One Complex Variable
+grand_parent: Complex Analysis
+nav_order: 3
+# cspell:words
+---
+
+# {{ page.title }}
+
+{: .theorem-title }
+> {{ page.title }}
+>
+> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
+> Let $\gamma$ be a contour in $G$ such that the interior of $\gamma$ is contained in $G$.
+> Then for any point $a$ in the interior of $\gamma$,
+>
+> $$
+> f(a) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} \, dz.
+> $$
+> {: .katex-display .mb-0 }
+
+{% proof %}
+{% endproof %}
+
+{: .theorem-title }
+> {{ page.title }} (Generalization)
+> {: #cauchys-integral-formula-generalized }
+>
+> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
+> Then the $n$th derivative $f^{(n)}$ exists for every $n \in \NN$.
+> If $\gamma$ is a contour in $G$ such that the interior of $\gamma$ is contained in $G$,
+> then for any point $a$ in the interior of $\gamma$,
+>
+> $$
+> f^{(n)} (a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz.
+> $$
+> {: .katex-display .mb-0 }
+
+{% proof %}
+{% endproof %}
+
+The last formula may be rewritten as
+
+$$
+\int_{\gamma} \frac{f(z)}{(z-a)^n} \, dz = \frac{2 \pi i}{(n-1)!} f^{(n-1)}(a)
+$$
+
+and is often used to compute the integral.
+
+## Many Consequences
+
+{: .theorem-title }
+> Cauchy's Estimate 
+> {: #cauchys-estimate }
+>
+> Let $f$ be holomorphic on an open set containing the disc with center $a$ and radius $r>0$.
+> Then
+>
+> $$
+> \norm{f^{(n)}(a)} \le \frac{n!}{r^n} \sup_{\abs{z-a} = r} \norm{f(z)} \qquad \forall n \in \NN.
+> $$
+> {: .katex-display .mb-0 }
+
+{% proof %}
+From [{{ page.title }}](#cauchys-integral-formula-generalized)
+for the circular contour around $a$ with radius $r$ we obtain
+
+$$
+\begin{aligned}
+\norm{f^{(n)}(a)} &\le \frac{n!}{2\pi} \sup_{\abs{z-a} = r} \norm{f(z)} \, \int_{\abs{z-a} = r} \frac{dz}{\abs{z-a}^{n+1}}.
+\end{aligned}
+$$
+
+Note that the supremum is finite (and is attained),
+because $f$ is continuous and the circle is compact.
+Clearly, the integral evaluates to $2 \pi r / r^{n+1}$
+and the right hand side of the inequality reduces to the desired expression.
+{% endproof %}
+
+---
+
+Recall that an *entire* function is a holomorphic function that is defined everywhere in the complex plane.
+
+{: .theorem-title }
+> Liouville's Theorem
+> {: #liouvilles-theorem }
+>
+> Every bounded entire function is constant.
+
+{% proof %}
+Consider an entire function $f$ and assume that $\norm{f(z)} \le M$ for all $z \in \CC$ and some $M > 0$.
+Since $f$ is holomorphic on the whole plane, we may make
+[Cauchy's Estimate](#cauchys-estimate)
+for all disks centered at any point $a \in \CC$ and with any radius $r>0$.
+For the first derivative, we have $\norm{f'(a)} \le M/r$, which tends to $0$ for $r \to \infty$.
+Hence $f' = 0$ in the whole plane. This
+[implies](/pages/complex-analysis/one-complex-variable/basics.html#holomorphic-function-is-constant-if-derivative-vanishes)
+that $f$ is constant.
+{% endproof %}
+
+---
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