From 61b48fd8baae321daf294ebf670ef4906240d260 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Wed, 5 May 2021 19:03:09 -0400
Subject: homeworks and attendance quizzes for kang

---
 kang/att3.tex | 24 ++++++++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 kang/att3.tex

(limited to 'kang/att3.tex')

diff --git a/kang/att3.tex b/kang/att3.tex
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+++ b/kang/att3.tex
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+\noindent 1. Over $|z| = 1,$ find the integral of $e^{2z}/z^4$
+
+Given $f(z) = e^{2z},$ $\int_C {f(z) dz\over z^4} = {2\pi i\over 3!}
+f^{(3)}(0) = {8\pi i\over 3}$
+
+\noindent 2. Prove theorem.
+
+$|z-z_0|$ is, on $C_R,$ exactly $R.$ Therefore, the upper bound of the
+absolute value of the nth derivative of $f$ at $z_0$ becomes
+$${n! \over 2*\pi*i R^{n+1}} |\int_C f(z)dz|.$$
+
+$$|\int_C f(z) dz| < \int_C |f(z)| dz \leq 2\pi RM_R,$$
+so $$|f^{(n)}(z_0)| \leq {n!M_R \over R^n}.$$
+
+\noindent 3. If f is entire and bounded, it is a constant
+
+$|f(z)| <= M_R$, where $M_R$ is some number.
+
+This means the theorem from question 2 applies, and because $R$ can be
+arbitrarily large, the first derivative at $z_0$ (any point in the
+plane) must be bounded by $n!M_R/R \to 0$ as $R\to\infty$. This means
+the function is constant.
+
+\bye
-- 
cgit 

