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/hw6.tex | 64 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 64 insertions(+)
 create mode 100644 kang/hw6.tex

(limited to 'kang/hw6.tex')

diff --git a/kang/hw6.tex b/kang/hw6.tex
new file mode 100644
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+++ b/kang/hw6.tex
@@ -0,0 +1,64 @@
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\def\Log{\mathop{\rm Log}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+% page 132
+\noindent{\bf 1.}
+
+\noindent{\it (a)}
+
+$$\int_0^\pi {2e^{i\theta}+2\over 2e^{i\theta}} 2ie^{i\theta} d\theta
+= \int_0^\pi i(2e^{i\theta} + 2) d\theta
+= \int_0^\pi 2ie^{i\theta} d\theta + \int_0^\pi 2i d\theta
+= 2(-1 - 1) + 2\pi i.$$
+
+\noindent{\it (b)}
+
+$$\int_\pi^{2\pi} {2e^{i\theta} + 2\over 2e^{i\theta}} 2ie^{i\theta} d\theta
+= \int_\pi^{2\pi} i(2e^{i\theta} + 2) d\theta
+= \int_\pi^{2\pi} 2ie^{i\theta} d\theta + \int_\pi^{2\pi} 2 d\theta
+= 4 + 2\pi i.
+$$
+
+\noindent{\it (c)}
+
+This is equivalent to the sum of integrals over $(0, \pi)$ and $(\pi,
+2\pi).$
+
+\noindent{\bf 4.}
+
+Parameterizing the curve as $x + ix^3,$ with $x\in (-1, 1),$
+this integral can be taken as the two integrals (and the discontinuity
+ignored because the function is defined at either side of it),
+$$\int_{-1}^0 (1+3x^2) dx + \int_0^1 4x^3 (1+3x^2) dx
+= (1 + 1) + (1 + 2) = 5.$$
+
+\hrule
+%page 138
+\noindent{\bf 3.}
+
+The length of the triangle is $3+4+5 = 12,$ and the maximum value of
+$$e^z - \overline{z}$$ is $$|e^z - \overline{z}| \leq e^x + \sqrt{x^2 + y^2}
+\leq \max{e^z} + \max{\sqrt{x^2 + y^2}} \leq 1 + 4,$$
+so by the theorem, the integral will be less that $ML = 12*5 = 60.$
+
+\noindent{\bf 4.}
+
+The length of the top half of the circle will be $\pi R,$ and the
+maximum value of the integrand will be less than the quotient of the
+maximum value of the dividend and the minimum value of the divisor.
+$|2z^2 - 1|$ has a maximum value of $2R^2 + 1$ at $z = iR,$ and
+$$z^4 + 5z^2 + 4 = (z^2 + 1)(z^2 + 4)$$ has a minimum value of $(R^2 -
+1)(R^2 - 4)$ (coincidentally at $z = iR,$ although this isn't necessary)
+
+The upper bound for this integral is, therefore, $ML,$ which corresponds
+to the given upper bound.
+
+Dividing the numerator and the denominator by $R^4$ gives
+$$\pi({1/R} + O(1/R^2)) \over 1 + O(1/R),$$ and because $1/R$ and
+$1/R^2$ both approach $0$ as $R$ approaches $\infty,$ we get
+$$\pi(1/R),$$ which also approaches $0.$
+
+\bye
-- 
cgit