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/hw10.tex | 153 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 153 insertions(+)
 create mode 100644 kang/hw10.tex

(limited to 'kang/hw10.tex')

diff --git a/kang/hw10.tex b/kang/hw10.tex
new file mode 100644
index 0000000..198067a
--- /dev/null
+++ b/kang/hw10.tex
@@ -0,0 +1,153 @@
+\def\Res{\mathop{\rm Res}}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 237
+\noindent{\bf 3.}
+
+With
+$$f(z) = {4z-5\over z(z-1)},$$
+
+$$\int_C f(z)dz = -2\pi i\Res_{z=\infty}f(z) =
+2\pi i\Res_{z=0}{1\over z^2}f\left({1\over z}\right) =
+2\pi i\Res_{z=0}{z(4 - 5z) \over z^2(1 - z)}.$$
+
+$${z(4-5z)\over z^2(1-z)} = (4/z-5)(1+z+z^2+\cdots) \to b_1 = 4.$$
+Therefore, the value of the integral is $8\pi i.$
+
+\hrule
+%page 242
+\noindent{\bf 1.}
+
+\noindent{\it (a)}
+
+$$f(z) = z\exp\left({1\over z}\right) = z\left(1 + {1\over z} +
+{1\over 2z^2} + \cdots\right),$$
+giving principal part
+$${1\over 2z} + {1\over 6z^2} + \cdots,$$
+meaning that $z=0$ is an essential singular point.
+
+\noindent{\it (b)}
+
+$$f(z) = {z^2\over 1+z} = {z^2\over 1 - (-z)} = z^2 - z + 1 - {1\over z}
++ {1\over z^2} + \cdots,$$
+giving principal part
+$$-{1\over z} + {1\over z^2} + \cdots,$$
+demonstrating an essential singular point at $z=0.$
+
+\noindent{\it (c)}
+
+$$f(z) = {\sin z\over z} = 1 - {z^2\over 3!} + {z^4\over 5!} - {z^6\over
+7!}.$$
+The principal part is $0,$ so $z=0$ is a removable singular point.
+
+\noindent{\it (d)}
+
+$$f(z) = {\cos z\over z} = {1\over z} - {z\over 2!} + {z^3\over 4!} +
+\cdots,$$
+giving principal part $1/z,$ meaning $z=0$ is a simple pole.
+
+\noindent{\it (e)}
+
+$$f(z) = {1\over (2-z)^3} = -{1\over (z-2)^3}$$
+is a complete Laurent series, and also its principal part, telling us
+there is a pole of order 3 at $z=2.$
+
+\noindent{\bf 3.}
+
+\noindent{\it (a)}
+
+If $f(z_0) \neq 0,$ $f(z)$ has Taylor series representation $a_0 +
+{a_1(z-z_0)\over 1!} + {a_2(z-z_0)^2\over 2!} + {a_3(z-z_0)^3\over 3!}
++ \cdots.$ Therefore, $g(z)$ has Laurent series representation
+${a_0\over z-z_0} + a_1 + {a_2(z-z_0)\over 2!} + {a_3(z-z_0)^3\over 3!},$
+evidencing a simple pole.
+
+\noindent{\it (b)}
+
+Similarly, if $f(z_0) = 0,$ $g(z)$ has Taylor series ${a_1\over 1!} +
+{a_2(z-z_0)\over 2!} + {a_3(z-z_0)^2\over 3!},$ with principal part $0$
+and a removable singular point.
+
+\hrule
+%page 246
+\noindent{\bf 3.}
+
+\noindent{\it (a)}
+
+$\sinh z/z = f(z)$ is non-zero and analytic at $z=0,$ meaning that
+$g(z) = \sinh z/z^4 = f(z)/z^3$ has a pole of order 3 at $z=0.$
+The Laurent series of $\sinh z/z^4$ is ${1\over z^3} + {1\over 3!z} +
+{z\over 5!} + \cdots,$ giving a residue of $1/6.$
+
+\noindent{\it (b)}
+
+$${1\over z(e^z - 1)} = {1\over z(1+z+{z^2\over 2}+{z^3\over
+6}+\cdots-1)} = {1\over z^2(1 + {z\over 2} + {z^2\over 6} + \cdots)} =
+{1\over z^2}{1\over 1 - (-{z\over 2} - {z^2\over 6} - \cdots)},$$
+and by the standard $1/(1-z)$ expansion since this series is less than 1
+(it equals ${e^z - 1\over z}-1,$ which is near 0 about $z=0,$ which can
+be verified by L'hospital's rule),
+$$ = {1\over z^2} - {1\over 2z} - {1\over 6} + \cdots$$
+
+This is a residue of $-1/2.$
+
+\noindent{\bf 5.}
+
+\noindent{\it (a)}
+
+$|z|=2$ contains the singular point at $z=0,$ and $f(z)$ can be defined
+as
+$$\phi(z)\over z^3$$ where $\phi(z) = 1/(z+4),$ which is analytic within
+the region. The residue is $\phi''(0)/2! = {2/(0+4)^3\over 2} = 1/64.$
+This gives the integral a value of $2\pi i/64 = \pi i/32.$
+
+\noindent{\it (b)}
+
+This region contains both singularities at $z=0$ and $z=-4,$ and its
+residue is the sum of theirs.
+We have already evaluated the residue at $z=0,$ so we will now evaluate
+the residue at $z=-4.$
+$$f(z) = {{1\over z^3}\over z+4} = {\phi(z)\over z+4}.$$
+This has residue $\phi(-4) = -1/64.$
+With the residue at $z=0,$ this gives a sum total of $0.$
+
+\noindent{\bf 6.}
+
+$C$ contains singularities at $-i,$ $0,$ and $i.$
+At $z=0,$
+$$f(z) = {\phi(z)\over z},$$
+with
+$$\phi(z) = {\cosh \pi z\over z^2+1}.$$
+This tells us the residue is $\phi(0) = \cosh 0/1 = 1.$
+
+Similarly, $z=i$ has
+$$f(z) = {\phi(z)\over z-i},$$
+with
+$$\phi(z) = {\cosh \pi z\over z(z+i)} \to \phi(i) =
+{\cosh(\pi i)\over -2} = {1\over 2}.$$
+
+And $z=-i$ has
+$$f(z) = {\phi(z)\over z+i},$$
+with
+$$\phi(z) = {\cosh \pi z\over z(z-i)} \to \phi(i) =
+{\cosh(-\pi i)\over 2} = {1\over 2}.$$
+
+The sum residue is $2,$ giving the integral a value of $2\cdot 2\pi i =
+4\pi i.$
+
+\iffalse
+$C$ contains all singularities, so the integral can be determined as
+$2\pi i\Res_{z=\infty} f(z).$
+
+$${1\over z^2}f\left({1\over z}\right) =
+{z^3\cosh(\pi/z) \over 1+z^2} =
+{z^3\over 1+z^2}\left(1 + {(\pi/z)^2\over 2!} + {(\pi/z)^4\over
+4!}\right) =
+
+%{1\over 1+z^2}\left(z^3 + {z\pi^2\over 2!} + {\pi^4\over 4!z} +
+%{\pi^6\over 6!z^3} + \cdots\right)
+$$
+\fi
+
+\bye
-- 
cgit 

