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/att1     |  33 +++++++++++
 kang/att1.png | Bin 0 -> 24029 bytes
 kang/att2     |  14 +++++
 kang/att3.tex |  24 ++++++++
 kang/att4     |  19 ++++++
 kang/att5     |   3 +
 kang/att5.tex |  32 ++++++++++
 kang/att6.tex |  25 ++++++++
 kang/hw10.tex | 153 ++++++++++++++++++++++++++++++++++++++++++++++++
 kang/hw11.tex | 175 ++++++++++++++++++++++++++++++++++++++++++++++++++++++
 kang/hw12.tex |  79 +++++++++++++++++++++++++
 kang/hw13.tex | 126 +++++++++++++++++++++++++++++++++++++++
 kang/hw14.tex | 103 ++++++++++++++++++++++++++++++++
 kang/hw2.tex  | 166 ++++++++++++++++++++++++++++++++++++++++++++++++++++
 kang/hw3.tex  |  84 ++++++++++++++++++++++++++
 kang/hw4.tex  | 163 +++++++++++++++++++++++++++++++++++++++++++++++++++
 kang/hw5.tex  | 185 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 kang/hw6.tex  |  64 ++++++++++++++++++++
 kang/hw7.tex  |  92 +++++++++++++++++++++++++++++
 kang/hw8.tex  |  98 +++++++++++++++++++++++++++++++
 kang/hw9.tex  | 185 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 21 files changed, 1823 insertions(+)
 create mode 100644 kang/att1
 create mode 100644 kang/att1.png
 create mode 100644 kang/att2
 create mode 100644 kang/att3.tex
 create mode 100644 kang/att4
 create mode 100644 kang/att5
 create mode 100644 kang/att5.tex
 create mode 100644 kang/att6.tex
 create mode 100644 kang/hw10.tex
 create mode 100644 kang/hw11.tex
 create mode 100644 kang/hw12.tex
 create mode 100644 kang/hw13.tex
 create mode 100644 kang/hw14.tex
 create mode 100644 kang/hw2.tex
 create mode 100644 kang/hw3.tex
 create mode 100644 kang/hw4.tex
 create mode 100644 kang/hw5.tex
 create mode 100644 kang/hw6.tex
 create mode 100644 kang/hw7.tex
 create mode 100644 kang/hw8.tex
 create mode 100644 kang/hw9.tex

diff --git a/kang/att1 b/kang/att1
new file mode 100644
index 0000000..bc3ea17
--- /dev/null
+++ b/kang/att1
@@ -0,0 +1,33 @@
+1)
+
+(z^2+3)/((z+1)(z^2+5))
+
+Product rule, chain rule, and power rule can be used to find a
+derivative except where the function is undefined, (z+1)(z^2+5) = 0.
+These are z = -1, \pm\sqrt{5}
+
+2)
+
+f(x+iy) = xy + iy
+
+Ux = y
+Uy = x
+Vx = 0
+Vy = i
+
+Ux = Vy? y is a real number, so it is not equal to i. This is,
+therefore, not analytic.
+
+3)
+
+f(z) = z^2 = x^2 + 2xyi - y^2
+
+Ux = 2x
+Uy = -2y
+Vx = 2y
+Vy = 2x
+
+Ux = Vy? Yes.
+Uy = -Vx? Yes.
+
+This function is analytic everywhere on the plane.
diff --git a/kang/att1.png b/kang/att1.png
new file mode 100644
index 0000000..f591ac1
Binary files /dev/null and b/kang/att1.png differ
diff --git a/kang/att2 b/kang/att2
new file mode 100644
index 0000000..25dc428
--- /dev/null
+++ b/kang/att2
@@ -0,0 +1,14 @@
+Semicircle of sqrt(z)
+value
+
+i 2^2 / {3/2 i} ( e^{3/2 i pi} - 1 )
+
+
+z + 1
+------
+(z^2+4)(z^2+9)
+upper bound
+
+(r+1) pi r
+------
+|r^2-1||r^2-1|
diff --git a/kang/att3.tex b/kang/att3.tex
new file mode 100644
index 0000000..2afdb9a
--- /dev/null
+++ b/kang/att3.tex
@@ -0,0 +1,24 @@
+\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
diff --git a/kang/att4 b/kang/att4
new file mode 100644
index 0000000..8c19a35
--- /dev/null
+++ b/kang/att4
@@ -0,0 +1,19 @@
+Q1) Find power series
+
+f(z) = e^{-z}/z^2 for 0<|z|<\infty
+
+e^{-z} = 1 - z + z^2/2! - z^3/3! + z^4/4! ...
+f(z) = 1/z^2 - 1/z + 1/2! - z/3! + z^2/4! ...
+
+Q2)
+
+f(z) = {1+2z^2 \over z^3 + z^5} |z|<1
+     = {1 \over z^3(1+z^2)} + {2\over z(1+z^2)}
+
+{1\over (1+z^2)} = 1 - z^2 + z^4 - z^6 ...
+
+{1\over z^3(1+z^2)} = 1/z^3 - 1/z + z - z^3 ...
+{2\over z(1+z^2)} = 2/z - 2z + 2z^3 - 2z^5 ...
+
+f(z) =
+1/z^3 + 2/z - 1/z - 2z + z + 2z^3 - z^3 - 2z^5 ...
diff --git a/kang/att5 b/kang/att5
new file mode 100644
index 0000000..867b76a
--- /dev/null
+++ b/kang/att5
@@ -0,0 +1,3 @@
+- favorite fruit and greek letter
+
+- how is life
diff --git a/kang/att5.tex b/kang/att5.tex
new file mode 100644
index 0000000..7c0cc02
--- /dev/null
+++ b/kang/att5.tex
@@ -0,0 +1,32 @@
+Q1)
+
+$$f(z) = {4z-5\over z(z-1)}.$$
+
+To determine the residue of $|z|=2,$ we sum the residues of interior
+poles ($z=0$ and $z=1.$)
+
+Starting with $z=0,$
+$$f(z) = -{4z-5\over z}\cdot{1\over 1-z} = (5/z-4)(1+z+z^2+\ldots),$$
+giving a residue of $b_1 = 5.$
+
+Then, $z=1,$
+$$f(z) = {4(z-1)-1\over z-1}\cdot{1\over 1-(1-z)} = (4 -
+1/(z-1))(1-(z-1)+(z-1)^2+\ldots),$$
+giving a residue of $b_1 = -1.$
+
+The sum of these residues is $4,$ so the value of the given integral is
+$2\pi i\cdot4 = 8\pi i.$
+
+Q2)
+
+$$f(z) = {z^3 + 2z\over (z-i)^3}.$$
+
+The residue at $z=i$ is $g''(i)/2!$ where $g(z) = z^3 + 2z,$ giving
+$g''(i)/2! = (6i)/2! = 3i.$
+
+Q3)
+
+I'm going to sleep in, and I think I'll be able to actually relax on
+Wednesday.
+
+\bye
diff --git a/kang/att6.tex b/kang/att6.tex
new file mode 100644
index 0000000..592ccc0
--- /dev/null
+++ b/kang/att6.tex
@@ -0,0 +1,25 @@
+\noindent{\bf Q1)}
+
+$$w = {i-z\over i+z} = -{z-i\over z+i} = e^{i\pi}{z-i\over z-\overline i}
+.$$
+
+This, therefore, puts the upper half-plane into a circle of radius 1.
+
+\noindent{\bf Q2)}
+
+$$w = {z-1\over z+1} = {x+iy-1\over x+iy+1} = {(x-1+iy)(x+1-iy)\over
+(x+1)^2+y^2} = {x^2 - 1 + 2iy + y^2\over (x+1)^2+y^2}.$$
+
+$$u = {x^2 - 1 + y^2\over (x+1)^2 + y^2},\quad v = {2y\over(x+1)^2+y^2}
+.$$
+
+$$y > 0 \to v > 0,$$
+because, with $x = -1,$ $v = 2/y,$ which maps positive y to positive v.
+
+\noindent{\bf Q3)}
+
+They are different because the transformation in Q2 isn't the same as
+Q1. It could be rewritten as a rotation of $pi/2$ then a transform
+similar to Q1, but this does not correspond.
+
+\bye
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
diff --git a/kang/hw11.tex b/kang/hw11.tex
new file mode 100644
index 0000000..d55a9e8
--- /dev/null
+++ b/kang/hw11.tex
@@ -0,0 +1,175 @@
+\def\Res{\mathop{\rm Res}}
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\let\rule\hrule
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+%page 253
+
+\noindent{\bf 4.}
+
+\noindent{\it (a)}
+
+$$(\sec z)^{-1} = \cos z = q(z).$$
+$$p(z) = z.$$
+$$p(\pi/2+n\pi) = \pi/2 + n\pi = z_n\neq 0, \qquad q(\pi/2+n\pi) = 0,
+\qquad q'(\pi/2+n\pi) = -\sin(\pi/2+n\pi) = (-1)^{n+1} \neq 0,$$
+so this is a simple pole with residue
+$$\Res_{z=z_n}(z\sec z) = (-1)^nz_n.$$
+
+\noindent{\it (b)}
+
+Similarly,
+$$(\tanh(z)) = {\sinh z\over\cosh z},$$
+giving $p(z) = \sinh z$ and $q(z) = \cosh z,$
+which forms a simple pole because
+$$p(z_n) = -\sin(\pi/2 + n\pi) = (-1)^n, \qquad q(z_n) = \cos(\pi/2 +
+n\pi) = 0, \qquad q'(z_n) = \sin(\pi/2 + n\pi) = (-1)^n.$$
+
+$$p(z_n)/q'(z_n) = (-1)^n/(-1)^n = 1 = \Res_{z=z_n}\tanh z$$
+
+\noindent{\bf 6.}
+
+The value of this integral is equal to $2\pi i$ times the sum of the
+residues of $$f(z) = {1\over z^2\sin z}$$ within the square.
+These are $z^2 = 0 \to z = 0$ and $\sin z = 0 \to z = z_n = \pi n,$
+where $-N\leq n\leq N.$
+This set includes $z = 0.$
+
+$$f(z) = {p(z)\over q(z)}$$
+where $p(z) = 1$ and $q(z) = z^2\sin z.$ $p(z_n)\neq 1,$ $q(z_n) = 0,$
+and $q(z_n) = z_n^2\cos z_n + 2z_n\sin z_n = z_n^2(-1)^n \neq 0$ (except
+at $z_n = 0.$)
+The sum of residues within the range of $-N\leq n\leq N$ and $n\neq 0$
+is $$\sum_{n=1}^N (-1)^nz_n^{-2}+z_{-n}^{-2} = \sum_{n=1}^N {2(-1)^n\over(\pi
+n)^2}.$$
+
+At $z_n = 0,$
+$$f(z) = {\phi(z)\over z^3},$$
+where $\phi(z) = {z\over\sin z},$
+This gives a residue at 0 of ${\phi''(0)\over 3!}$.
+$$\phi''(z) = -{\cos z\over(\sin z)^2} - {\cos z - z\sin z\over(\sin z)^2}
++ {2z(\cos z)^2\over(\sin z)^3} =
+{-2\cos z\sin z + z(\sin z)^2 + 2z(\cos z)^2 \over (\sin z)^3}.$$
+$$\phi''(0) = 1,$$
+found by noting that, near 0, $\sin z\to z$ and $\cos z\to 1.$
+This gives a 1/6 residue.
+
+Together, this makes
+$$\int_{C_N} {dz\over z^2\sin z} = 2\pi i\left[{1\over 6}+2\sum_{n=1}^N
+{(-1)^n\over\pi^2 n^2}\right].$$
+
+If the integral tends to 0 as $N\to\infty,$
+$$0 = {1\over 6}+2\sum_{n=1}^N{(-1)^n\over \pi^2 n^2} \to 
+-{\pi^2\over 12} = \sum_{n=1}^N{(-1)^n\over n^2},$$
+meaning the series tends to $\pi^2/12.$
+
+\noindent{\bf 10.}
+
+Because $p(z)/q(z)$ has a pole of order $m,$
+$${p(z)\over q(z)} = {\phi(z)\over (z-z_0)^m},$$
+where $\phi(z)$ is analytic and non-zero at $z_0,$ further giving
+$$q(z) = {p(z)\over\phi(z)}(z-z_0)^m,$$
+and because $\phi(z)$ is nonzero, $p(z)/\phi(z)$ is analytic, and this
+tells us that $q(z)$ has a zero of order $m$ because $q(z) =
+g(z)(z-z_0)^m$ where $g(z)$ is a nonzero analytic function.
+
+\hrule
+%page 264
+
+\noindent{\bf 1.}
+
+$$z^2+1 = (z-i)(z+i).$$
+The only singularity above the real axis is $z = i.$
+This point is a simple pole, so its residue is $1/(i+i) = -i/2.$
+This gives the value of the integral as $\pi i*(-i/2) = \pi/2$ (since
+the integrand is even and $\pi R/(R^2-1)$ vanishes to 0 as $R$ tends to
+infinity).
+
+\noindent{\bf 4.}
+
+The singularities of this integrand are, like in the example, the sixth
+roots of $-1,$ and they are simple poles, but the residues $B_k$
+evaluate instead to $$B_k = \Res_{z=c_k} {z^2\over z^6+1} = {c_k^2\over
+6c_k^5} = {c_k^3\over 6c_k^6} = -{c_k^3\over 6}.$$
+$$c_k = e^{i(\pi/6+k\pi/3)} \to c_k^3 = e^{i(\pi/2+k\pi)} = i(-1)^k.$$
+Since $k\in\{0,1,2\},$ the residues sum to $-i/6,$ giving the value of
+the integral to be $\pi i*(-i/6) = -\pi/6,$ (again valid because the
+integrand is even and $\pi R^3/(R^6-1)$ vanishes to 0 as $R$ tends to
+infinity).
+
+\noindent{\bf 8.}
+
+$$q(z)=(z^2+1)(z^2+2z+2) = (z+i)(z-i)(z+(1+i))(z+(1-i)).$$
+The singularities above the real axis are simple poles $z = i, -1+i.$
+At these singularities, the residues $p(z)/q'(z)$ are respectively
+$${i\over (2i)(1+2i)(1)} = {1-2i\over10}$$ and $${-1+i\over
+(-1+2i)(-1)(2i)} = {-1+3i\over 10}.$$
+
+$M_R = {R\over (R^2-1)(R^2-2)} \to R\pi M_R = {R^2\over
+(R^2-1)(R^2-2)},$
+so $\int_{C_R}f(x)dx = 0$ as $R$ tends to infinity.
+This allows us to determine the Cauchy principal value of the integral
+to be $2\pi i\cdot i/10 = -\pi/5.$
+
+\hrule
+%page 273
+
+\noindent{\bf 1.}
+
+$$\int_{-\infty}^\infty {\cos x\thinspace dx\over (x^2+a^2)(x^2+b^2)} =
+\Re \int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)}.$$
+The singularities in the upper half of the plane are simple poles
+$x=ai, x=bi.$
+These have residues $${e^{-a}\over (2ai)(-a^2+b^2)}$$ and $$e^{-b}\over
+(2bi)(-b^2+a^2),$$ respectively.
+$$\int_{-\infty}^\infty {e^{ix}dx\over (x^2+a^2)(x^2+b^2)}
+= 2\pi iB - \int_{C_R} f(z)e^{iz} dz.$$
+This last integral tends to $0$ by Jordan's lemma because $f(z)$ has
+maximum value ${1\over (R^2-a^2)(R^2-b^2)},$ which tends to 0 as $R$
+tends to infinity where $R>a$ and $R>b.$
+$$2\pi iB = 2\pi i\left({e^{-a}/2a-e^{-b}/2b \over i(a^2-b^2)}\right) =
+\pi({e^{-a}/a - e^{-b}/b \over (a^2-b^2)}),$$
+proving the integral formula.
+
+\noindent{\bf 5.}
+
+$$\int_{-\infty}^\infty {x^3\sin ax\over x^4+4}dx = \Im
+\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx.$$
+
+$$\int_{-\infty}^\infty {x^3e^{iax}\over x^4+4}dx = 2\pi iB - \int_{C_R}
+f(x)e^{iax}dx.$$
+Where $|x|>R>\sqrt 2,$ the maximum value of $f(x)$ is
+$$M_R<{R^3\over R^4-4},$$
+which tends to $0$ as $R\to\infty,$ telling us the integral evaluates to
+$0.$
+The singularities in the upper half of the plane are $c_k = \sqrt
+2e^{i(\pi/4+k\pi/2)},$ where $k\in\{0,1\}.$
+These are simple poles, so they evaluate to $$p(c_k)/q'(c_k) = {c_k^3
+e^{iac_k}\over 4c_k^3} = e^{iac_k}/4 =
+e^{ia\sqrt2 e^{i(\pi/4+k\pi/2)}}/4 =
+$$$$ e^{ia\sqrt2(\cos(\pi/4+k\pi/2)+i\sin(\pi/4+k\pi/2))}/4
+= e^{ia((-1)^k+i)}/4
+= e^{a(i(-1)^k-1)}/4
+= e^{-a}(\cos(a(-1)^k)+i\sin(a(-1)^k))/4,$$
+summing to $2e^{-a}\cos(a)/4.$
+Therefore, the integral evaluates to $2\pi ie^{-a}\cos a/2,$
+giving imaginary component (and value of the original integral)
+$\pi e^{-a}\cos a.$
+
+\noindent{\bf 8.}
+
+$$\int_{-\infty}^\infty {\sin x dx\over x^2+4x+5} =
+\Im\int_{-\infty}^\infty {e^{ix} dx\over (x+(2+i))(x+(2-i))}$$
+
+The only singularity in the upper half of the plane is $-2+i,$
+with a residue of
+$$e^{-2i-1}\over 2i,$$
+giving the integral a value
+${2\pi ie^{-2i-1}\over 2i} = \pi e^{-1-2i} = {\pi\over e}(\cos(-2)+i\sin(-2)),$
+giving an imaginary component $-\pi\sin(2)/e.$
+
+This answer is valid because $f(z)$ tends to zero on $C_R$ as
+$R\to\infty,$ where $$f(z) = {1\over z^2+4z+5}.$$
+
+\bye
diff --git a/kang/hw12.tex b/kang/hw12.tex
new file mode 100644
index 0000000..041c817
--- /dev/null
+++ b/kang/hw12.tex
@@ -0,0 +1,79 @@
+\def\Re{\mathop{Re}\nolimits}
+\def\Res{\mathop{Res}}
+\let\rule\hrule
+\def\fr#1#2{{#1\over#2}}
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+% page 282
+\noindent{\bf 1.}
+
+With
+$$f(z) = {e^{iaz} - e^{ibz}\over z^2},$$
+there is one singularity at 0, so by Cauchy-Goursat,
+$$\int_{C_\rho} f(z)dz + \int_{C_R} f(z)dz + \int_{-\infty}^0 f(z)dz +
+\int_0^\infty f(z)dz = 0,$$
+where $C_\rho$ is the upper semicircle about $z=0,$ of radius
+$\rho\to0,$ and similar for $C_R$ where $R\to\infty.$
+
+$$\int_{C_R} f(z)dz = \int_{C_R} {e^{iaz} - e^{ibz}\over z^2}dz
+= 0,$$
+by Jordan's lemma.
+$$\int_{C_\rho} {e^{iaz} - e^{ibz} \over z^2} dz = -B_0\pi i = \pi(a-b),$$
+by the theorem in this section.
+$$\int_{-\infty}^0 f(x)dx + \int_0^\infty f(x)dx =
+-\int_0^\infty f(-x)dx + \int_0^\infty f(x)dx =$$$$
+\int_0^\infty {e^{iax} - e^{ibx} + e^{-iax} - e^{-ibx}\over x^2}dx
+% logical discontinuity! It should maybe be - e^{-iax} + e^{-ibx}.
+= 2\int{\cos(ax)-\cos(bx)\over x^2}dx,$$
+which, by transformation of the original equality, gives
+$$2\int{\cos(ax)-\cos(bx)\over x^2}dx = -\pi(a-b) \to \int f(x)dx =
+{\pi\over2}(b-a).$$
+
+\noindent{\bf 4.}
+
+$$f(z) = {z^{1/3}\over(z+a)(z+b)} = {e^{(1/3)\log z}\over(z+a)(z+b)}.$$
+$$\int_\rho^R {r^{1/3}\over (r+a)(r+b)}dr +
+\int_{C_R} f(z)dz - \int_\rho^R {r^{1/3}e^{2i\pi/3}\over (r+a)(r+b)}dr +
+\int_{C_\rho} f(z)dz = 2\pi i(\Res_{z=-a}f(z) + \Res_{z=-b}f(z)).$$
+
+These residues are at simple poles and are therefore
+$${a^{1/3}e^{\pi i/3}\over b-a}\qquad{\rm and}\qquad
+{b^{1/3}e^{\pi i/3}\over a-b},$$
+respectively.
+
+$$\left|\int_{C_\rho} f(z)dz\right| \leq
+{\rho^{1/3}\over(a+\rho)(b+\rho)}2\pi\rho \to 0$$
+$$\left|\int_{C_R} f(z)dz\right| \leq {R^{1/3}\over (R+a)(R+b)}2\pi R\to
+0.$$
+
+Therefore, rearranging the original equality,
+$$(1-e^{2i\pi/3})\int_0^\infty {r^{1/3}\over (r+a)(r+b)}dr
+= 2\pi ie^{\pi i/3}{a^{1/3}-b^{1/3}\over b-a}$$
+$$\longrightarrow \int_0^\infty {x^{1/3}\over (x+a)(x+b)}dx =
+{2\pi\over\sqrt 3}\cdot{\root 3 \of a - \root 3 \of b\over b - a}$$
+
+\hrule
+% page 287
+
+\noindent{\bf 5.}
+
+$$\int_0^\pi {d\theta\over (a+\cos\theta)^2} = {1\over 2}\int_0^{2\pi}
+{d\theta\over (a+\cos\theta)^2},$$
+by symmetry of the integrand,
+$$=\fr12\int_C {1\over (a+{z+z^{-1}\over 2})^2}{dz\over iz} =
+\fr2i\int_C {z\over (z^2+2az+1)^2}.$$
+which has two singularities, those being
+$$z^2 + 2az + 1 = 0 \to z = \pm\sqrt{a^2 - 1} - a.$$
+Only the more positive singularity $z_0$ is within $|z|<1,$ where $a >
+1.$ The integrand can be rewritten as
+$${\phi(z)\over (z-z_0)^2}$$
+where $\phi(z) = {z\over (z-z_1)^2},$ $z_1$ being the more negative
+singularity.
+The singularity has residue
+$$B_0 = \phi'(z_0) = {-z_1-z_0\over (z_0-z_1)^3}
+= {2a\over 8\sqrt{a^2-1}^3},$$
+giving the integral value
+$$\int_0^\pi {d\theta\over (a+\cos\theta)^2} = 2\pi i\fr2i\cdot{2a\over
+8\sqrt{a^2-1}^3} = {a\pi\over\sqrt{a^2-1}^3}.$$
+
+\bye
diff --git a/kang/hw13.tex b/kang/hw13.tex
new file mode 100644
index 0000000..251a1d2
--- /dev/null
+++ b/kang/hw13.tex
@@ -0,0 +1,126 @@
+\def\arg{\mathop{arg}\nolimits}
+\let\rule\hrule
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+%page 293
+\noindent{\bf 2.}
+
+The graph of the contour image encircles the origin three times, giving
+$\Delta_C \arg f(z) = 2\pi\cdot 3 = 6\pi$ and, because it is analytic
+inside and on $C,$ $P = 0 \to 3 = Z - P = Z,$ or there are three zeroes,
+counting multiplicities, on $f(z)$ within $C.$
+
+\noindent{\bf 7.}
+
+\noindent{\it (a)}
+
+On $|z|=2,$
+$$36 = |9z^2| \geq 35 = 16 + 16 + 2 + 1 \geq |z^4 - 2z^3 + z - 1|,$$
+meaning the circle contains the same number of zeroes on the original
+and $9z^2,$ or 2, by Rouch\'e's theorem.
+
+\noindent{\it (b)}
+
+Similarly,
+$$32 = |z^5| \geq 29 = 24 + 4 + 1 \geq |3z^3+z^2+1|,$$
+giving 5 zeroes by the same method.
+
+\hrule
+%page 301
+\noindent{\bf 2.}
+
+$$w = iz + i = i(x+yi) + i = -y + (x+1)i = u+vi \to v = x+1,$$
+so if $x>0,$ $v>1.$
+
+\noindent{\bf 5.}
+
+$$w = (1-i)z = (1-i)(x+yi) = x - ix + yi + y = (x+y) + (y-x)i = u+vi,$$
+so by manipulating $u=x+y$ and $v=y-x,$
+$$y = (u+v)/2 > 1 \to u+v > 2 \leftrightarrow v > 2-u.$$
+
+\hrule
+%page 305
+\noindent{\bf 3.}
+
+$$y = {-v\over u^2+v^2} > c_2 \to c_2(u^2+v^2) + v < 0,$$
+which if $c_2>0$ becomes
+$$c_2u^2 + c_2(v+1/2c_2)^2 < 1/c_2^2,$$
+which is the interior of a circle.
+
+If $c_2 = 0,$ this is $v<0,$ or a half-plane, and if $c_2<0,$ the
+relation is inverted, and it is transformed to the exterior of a circle.
+
+\noindent{\bf 7.}
+
+This is first a leftward translation of $1$ and then the $w = 1/z$
+transform, or an inversion across $|z-1| = 1,$ then a reflection across
+the x-axis.
+
+\noindent{\bf 13.}
+
+With $z = r_0e^{i\theta},$
+
+$$w = z + {1\over z} = r_0e^{i\theta} + {1\over r_0}e^{-i\theta} =
+r_0(\cos\theta + i\sin\theta) + {1\over r_0}(\cos\theta - i\sin\theta) =
+u + vi$$
+
+$$u = (r_0+1/r_0)\cos\theta,\quad v = (r_0-1/r_0)\sin\theta.$$
+
+\hrule
+%page 311
+
+\noindent{\bf 3.}
+
+Since $0\mapsto\infty,$ $c\neq0$ and $d=0,$ giving
+$$w = {az+b\over cz}.$$
+$\infty\mapsto0,$ so $a=0,$ and $i\mapsto i$ gives
+$$i = {b\over ci} \to -c = b,$$
+thus
+$$w = {-c\over cz} = -{1\over z}$$
+
+\noindent{\bf 9.}
+
+If $f(0) = 0,$ and with $AB \neq 0,$
+$${w(w_2-w_3)\over (w-w_3)w_2} = {z(z_2-z_3)\over(z-z_3)z_2} \to
+Aw(z-z_3) = Bzw - Bzw_3 $$$$ w(A(z-z_3)-Bz) = - Bzw_3 \to w =
+{z\over (B-A)z/(Bw_3) - z_3/w_3},$$
+QED.
+
+\hrule
+%page 325
+
+\noindent{\bf 4.}
+
+$$w = e^z = e^{x+iy} = e^xe^{iy} = \rho e^{i\phi} \to \rho = e^x,\quad
+\phi = y.$$
+
+From the original constraints,
+$$0\leq\phi\leq\pi, \quad \rho\geq1.$$
+
+\def\li{\leavevmode\llap{\hbox to \parindent{\hfil$\bullet$\hfil}}}
+\noindent This gives boundaries:
+
+\li the inner semicircle of radius untiy, corresponding to the $x=0$
+boundary,
+
+\li the positive x-axis, corresponding to the $y=0$ boundary, and
+
+\li the negative x-axis, corresponding to the $y=\pi$ boundary.
+
+\noindent{\bf 9.}
+
+Using the following parametric representations,
+$$u = \sin x\cosh y,\quad v = \cos x\sinh y,$$
+the bottom of the rectangular region $y = a$ and $-\pi\leq x\leq\pi$
+becomes $u = \cosh a\sin x,$ $v = \sinh a\cos x,$ which is the
+parametrization of a complete ellipse with minor and major axes $\sinh
+a$ and $\cosh a,$ respectively.
+Similary, the upper bound parametrizes an ellipse with minor and major
+axes $\sinh b$ and $\cosh b.$
+This implies that the region between them continuously fills the space
+between them.
+
+The cut is between $x=\pi$ and $x=-\pi$ because $\sin\pi = \sin(-\pi)$
+and $\cos\pi=\cos(-\pi),$ meaning these boundaries overlap in the image.
+
+\bye
diff --git a/kang/hw14.tex b/kang/hw14.tex
new file mode 100644
index 0000000..4c21f9b
--- /dev/null
+++ b/kang/hw14.tex
@@ -0,0 +1,103 @@
+\def\arg{\mathop{\rm arg}\nolimits}
+\let\rule\hrule
+\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}
+
+%page 317
+\noindent{\bf 1.}
+
+$$w = {i-x\over i+x} = {(i-x)^2\over -1 - x^2} = -{-1 - 2ix + x^2 \over
+1+x^2} = {1-x^2\over 1+x^2} + i{2x\over 1+x^2}.$$
+$y=0,$ the boundary of the preimage in Figure 13 transforms into 
+
+$$|w| = {\sqrt{(1-x^2)^2+4x^2}\over 1+x^2} =
+{\sqrt{1-2x^2+x^4+4x^2}\over 1+x^2} = 1.$$
+
+\noindent{\bf 7.}
+
+Take $$\rho = \ln|Z|, \quad 0< \Theta = \arg(Z) \leq 2\pi.$$
+$$w = \rho + i\Theta.$$
+
+\hrule
+%page 330
+\noindent{\bf 1.}
+
+With constant $y_1,$
+$$u = x^2 - y_1^2, \quad v = 2xy_1.$$
+$$u = (v/2y_1)^2 - y_1^2 \to 4y_1^2(u+y_1^2) = v^2,$$
+assuming $y_1>0.$
+
+\noindent{\bf 4.}
+
+$y=0$ gives $u = x_1^2, v = 0,$ and $v(y) = -v(-y), u(y) = u(-y),$
+telling us that $x_1 = 1$ transforms into a complete parabolic region
+with form $v^2 = -4(u-1).$
+
+$x=\pm y$ similarly gives
+$$u = x^2 - x^2, \quad v = \pm 2x^2 \to u = 0, \quad v\in {\bf R,}$$
+which at the given boundary points has $A = 0 \mapsto A' = 0,$ $C =
+1\mapsto C' = 1,$ $D = 1+i \mapsto D' = 2,$ and $B' = 1-i\mapsto -2,$
+under the $w = z^2$ transformation.
+This completes the boundary of the region, which is interior to the
+boundary.
+
+\hrule
+%page 352
+\noindent{\bf 1.}
+
+$${dw\over dz} = 2z = 2(2+i)$$
+at $z_0 = 2+i,$ giving an angle of rotation $\arg 2+i = \tan^{-1}(1/2)$
+and scale factor $|2(2+i)| = 2\sqrt{5}.$
+This can be demonstrated with $y=1,$ passing through $z_0$ at angle
+$\theta = 0,$ transformed into
+$$(x+iy)^2 = (x+i)^2 = (t+2+i)^2 = t^2 + 2(1+i)t + 2,$$
+which, since $x(0) = z_0,$ gives an angle equal to the argument of
+$2(1+i),$ which is $\pi/4.$
+
+\noindent{\bf 4.}
+
+The transformation $w=z^n$ gives
+$${dw\over dz} = nz^{n-1},$$
+meaning that $\rho_0\exp(i\theta_0)$ gives angle of rotation
+$\arg(n\rho_0^{n-1}\exp(i(n-1)\theta_0)) = (n-1)\theta_0$ and scale
+factor $|n\rho_0^{n-1}\exp(i(n-1)\theta_0)| = n\rho_0^{n-1}.$
+
+\hrule
+%page 357
+\noindent{\bf 1.}
+
+\noindent{\it (a)}
+
+$$u(x, y) = 2x - x^3 + 3xy^2.$$
+
+This is harmonic because $$u_{xx} + u_{yy} = - 6x + 6x = 0$$ on all
+$x\in{\bf R}.$
+
+$$v_y = u_x = 2 + 3y^2 \to v(x,y) = 2y + y^3 + g(x).$$
+
+$$g'(x) = v_x = -u_y = -6xy = \to v(x,y) = 2y + y^3 - 3x^2y+C.$$
+
+\noindent{\it (b)}
+
+$$u(x,y) = \sinh x \sin y$$
+
+This is harmonic because $$u_{xx} + u_{yy} =
+\sinh x\sin y - \sinh x\sin y = 0.$$
+
+$$v_y = u_x = \cosh x\sin y \to v(x,y) = -\cosh x\cos y + g(x).$$
+$$g'(x) = v_x = -u_y = -\sinh x\cos y \to v(x,y) = -\cosh x\cos y -
+\cosh x\cos y + C = C - 2\cosh x\cos y.$$
+
+\noindent{\it (c)}
+
+$$u(x,y) = {y\over x^2+y^2}$$
+
+This is harmonic because $$u_{xx} + u_{yy} = 0.$$
+
+$$v_y = u_x = {-2yx \over (x^2+y^2)^2} \to v = {x\over x^2+y^2} + g(x).$$
+
+$$g'(x) = v_x = -u_y = -{x^2+y^2 - 2y^2\over (x^2+y^2)^2} \to g(x) =
+{x\over x^2+y^2} + C.$$
+
+% also read sec 127-128, 129 (question unnecessary)
+
+\bye
diff --git a/kang/hw2.tex b/kang/hw2.tex
new file mode 100644
index 0000000..5841038
--- /dev/null
+++ b/kang/hw2.tex
@@ -0,0 +1,166 @@
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\def\Arg{\mathop{\rm Arg}\nolimits}
+\def\sec#1{\vskip0pt plus .3fil\goodbreak\vskip 0pt plus -.3fil\medskip\noindent{\bf#1}}
+\tabskip=0pt plus 1fil
+
+% pg 13
+\sec{2.}
+
+Generally, $x<|x|,$ and $|z| = \sqrt{(\Re z)^2+(\Im z)^2}
+\to |z|^2 = |\Re z|^2 + |\Im z|^2 \to |\Re z|^2\leq |z|^2 \to |\Re z|
+\leq |z|.$
+
+\sec{5.}
+
+\halign to \hsize{&#\cr
+\vrule height 2in width 0pt\cr
+$|z-1+i=1|$&$|z+i|\leq3$&$|z-4i|\geq4$\cr}
+
+%circle centered at (1-i) of rad 1
+%closed circle centered at -i
+%everything but the circle of rad 4 around 4i, closed
+
+\sec{6.}
+
+$|z_1-z_2|$ is the distance between two points, so $|z-1| = |z+i|$
+represents every point $z$ that is equally distant from $(1,0)$ and
+$(0,-1).$ These points can be imagined as the intersection points of two
+expanding circles from these points. The expanding circles first meet at
+the midpoint, and then the intersection points move perpendicular to the
+line between them. The line between the two points has a slope of 1 and
+goes through $(.5,-.5),$ so the perpendicular line will go through the
+origin with slope $-1$.
+
+% pg 16
+\sec{2.}
+
+\halign to \hsize{&#\cr
+\vrule height 2in width 0pt\cr
+$\Re(\overline z - i) = 2$&$|2\overline z + i| = 4$\cr}
+
+%x = 2
+%|2\overline z + i| = |2z - i| = 2|z - i/2| = 4. Circ around i/2, rad 2
+
+\sec{7.}
+
+$|\overline z| = |z|,$ and because $|z|\leq 1,$ $|z^3|=|z|^3\leq 1.$
+
+$$|\Re(2+\overline z+z^3)| \leq |2+\overline z+z^3| \leq |2| +
+|\overline z| + |z^3| \leq 4,$$
+by addition of the maximum values of each value.
+
+\sec{14.}
+
+$x^2-y^2 = 1 \to (\Re z)^2-(\Im z)^2 = 1 =
+({z + \overline z\over 2})^2 - ({z - \overline z\over 2i})^2 =
+(1/4)(({z+\overline z})^2 + ({z-\overline z})^2) =
+(1/4)(z^2 + 2z\overline z + \overline z^2 + z^2 - 2z\overline z +
+\overline z^2) = (1/4)(2z^2 + 2\overline z^2)
+\to z^2 + \overline z^2 = 2.$
+
+% pg 23
+\sec{1.}
+
+\noindent{\it (a)}
+
+$$z = {-2\over 1+\sqrt 3 i}.$$
+
+$\arg z = \arg(-2) - \arg(1+\sqrt 3 i) = \pi - (\pi/3)+2\pi n = 2\pi/3 +
+2\pi n.$
+Therefore, $\Arg z = 2\pi/3$
+
+\noindent{\it (b)}
+
+$$z = \left(\sqrt3 - i\right)^6.$$
+
+$\arg z = 6\arg\left(\sqrt3 - i\right) = 6(-5\pi/6) = -5\pi
+\to \Arg z = \pi.$
+
+\sec{6.}
+
+If $\Re z > 0,$ $-\pi/2 < \Arg z < \pi/2$ because the absolute angle
+from the x-axis cannot be greater than $\pi/2.$
+
+Because $\arg(z_1z_2) = \arg z_1 + \arg z_2,$
+
+$$-\pi < \arg(z_1z_2) = \arg z_1 + \arg z_2 < \pi.$$
+
+% pg 30
+\sec{3.}
+
+$$|-8-8\sqrt 3 i| = 16.$$
+$$\arg(-8-8\sqrt 3 i) = -2\pi/3+2\pi n.$$
+$$(-8-8\sqrt 3 i)^{1/4} = 2(\cos(-\pi/6+n\pi/2)+i\sin(-\pi/6+n\pi/2)).$$
+Fully expanded, this is $\pm(\sqrt3-i)$ and $\pm(1+\sqrt3i),$ which form
+the vertices of a particular square. The principal root has argument
+$-\pi/6,$ so it is $\sqrt3-i$
+
+\sec{6.}
+
+$$z^4+4=0 \to z^4 = -4 \to z^4 = 4e^{i\pi+2n\pi} \to z = \sqrt2
+e^{i\pi/4 + n\pi/2},$$
+
+so the other zeroes are $\pm(-1+i)$ and $\pm(1+i).$
+
+$$(z-(1+i))(z-(1-i))(z+(1+i))(z+(1-i)) = (z^2-2z+2)(z^2+2z+2).$$
+
+%pg 34
+\sec{5.}
+
+$S_1,$ where $|z|<1$ and $S_2,$ where $|z-2|<1$ are both open sets, so
+they have no intersection. Because they don't intersect, there cannot be
+a polygonal line that is continuous and in both sets.
+
+\sec{7.}
+
+\noindent{\it (a)}
+
+$z_n = i^n$ has no accumulation points because the set
+$\{z_0,z_1,\ldots\} = \{1,i,-1,-i\},$ so any point $z$ other than these
+four would not contain them in its neighborhood $\eta = |z-z_0|,$ where
+$z_0$ is in the set (and these four points are not in their own deleted
+neighborhoods).
+
+\noindent{\it (b)}
+
+$0$ is the accumulation point of this set. $|z_n-0| = 1/n,$ so for all
+$\eta-{\rm neighborhoods}$ of $0$ include some elements of this set.
+
+\noindent{\it (c)}
+
+For this set, all points in $0\leq\arg z \leq \pi/2,$ and $z=0$ are
+accumulation points because that is the closure of this open, connected
+set.
+
+\noindent{\it (d)}
+
+For even $n,$ (i.e. $n$ where $(-1)^n = 1$) $|z_n - (1+i)| = 1/n,$ so
+$(1+i)$ is an accumulation point. Similar is true for odd $n$ and
+$-(1+i).$ No other points are accumulation points.
+
+%pg 43
+\sec{2.}
+
+\noindent{\it (a)}
+
+$$f(z) = z^3 + z + 1 = (x+iy)^3 + x + iy + 1 = x^3 + 3x^2yi - 3xy^2 -
+y^3i + x + iy + 1$$$$ = (x^3 - 3xy + x - 1) + i(3x^2y - y^3 + i).$$
+
+\noindent{\it (b)}
+
+$$f(z) = {\overline z^2\over z} = {\overline z^3\over z\overline z} =
+{\overline z^3\over |z|^2} = {x^3+3x^2yi-3xy^2+y^3i\over x^2+y^2} =
+{x^3-3xy^2\over x^2+y^2} + i{3x^2y+y^3\over x^2+y^2}.$$
+
+\sec{8.}
+
+Images of $r\leq1$ and $0\leq\theta\leq\pi/4.$
+
+\halign to \hsize{&#\cr
+\vrule height 2in width 0pt\cr
+$w=z^2$&$w=z^3$&$w=z^4$\cr}
+% r \leq 1 for all
+% 0\leq\theta\leq n\pi/4 where w=z^n
+
+\bye
diff --git a/kang/hw3.tex b/kang/hw3.tex
new file mode 100644
index 0000000..32cb710
--- /dev/null
+++ b/kang/hw3.tex
@@ -0,0 +1,84 @@
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+
+% page 54
+\noindent{\bf 7.}
+
+By the triangle theorem,
+$$|f(z)-w_0|+|w_0| \geq |f(z)| \to ||f(z)|-|w_0|| \leq
+|f(z)-w_0|<\delta,$$
+where $|z-z_0|<\epsilon$ by definition of the first limit.
+% not sure how to prove the abs transformation
+
+This shows that, as $z\to z_0,$ $|f(z)| \to |w_0|,$ because for all
+$\epsilon,$ there exists $\delta$ such that $||f(z)|-|w_0||\leq\delta.$
+
+\noindent{\bf 11.}
+
+\noindent{\it (a)}
+
+$\lim_{z\to\infty} T(z) = \infty$ is true iff $$\lim_{z\to 0}
+T(z^{-1})^{-1} = 0 = \lim_{z\to 0} {d\over az^{-1}+b} = \lim_{z\to 0}
+{d\over z^{-1}(a+bz)} = \lim_{z\to 0} {zd\over a+bz} =
+(\lim_{z\to 0} zd)/(\lim_{z\to 0} a+bz).$$
+
+$ad+bc \neq 0 \to ad \neq 0 \to a\neq0, d\neq0,$ so this becomes $0/a =
+0,$ and the theorem is proved.
+
+\noindent{\it (b)}
+
+When $c\neq 0$, $\lim_{z\to\infty} T(z) = a/c$ if
+$$\lim_{z\to 0} T(z^{-1}) = a/c = {az^{-1}+b\over cz^{-1}+d}
+= {z^{-1}(a+bz)\over z^{-1}(c+dz)} = {a+bz\over c+dz} = {\lim_{z\to 0}
+a+bz\over \lim_{z\to 0} c+dz} = a/c.$$
+
+$\lim_{z\to -d/c} T(z) = \infty$ if $$\lim_{z\to -d/c} {cz+d\over az+b}
+= 0 = {\lim_{z\to -d/c} cz+d\over\lim_{z\to -d/c} az+b} = {0\over k} = 0,$$
+
+and $k\neq 0$ because $ad-bc\neq0 \to -ad/c + b \neq 0.$
+
+% page 61
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+$f'(z) = (3z^2)' + (-2z)' + 4' = 6z - 2,$ by the addition rule of
+derivatives, $(af(x))' = af'(x),$ and the power rule.
+
+\noindent{\it (b)}
+
+By the chain rule, $g(w) = w^5,$ and $w = f(z) = 2z^2 + i,$ $$F'(z) =
+g'(w)f'(z) = 5w^4 \cdot 4z = 20(2z^2+i)^4z,$$
+with the power rule.
+
+\noindent{\it (c)}
+
+${z-1\over 2z+1} = (z-1)(2z+1)^{-1},$ so this can be solved with the
+product and chain rules:
+$${d\over dz} {z-1\over 2z+1} = (2z+1)^{-1} - {2(z-1)\over (2z+1)^2}.$$
+
+\noindent{\it (d)}
+
+With the product rule, and $f(z) = (1+z^2)^4$ and $g(z) = z^{-2},$
+$$F'(z) = f'(z)g(z) + f(z)g'(z)
+= {8z(1+z^2)^3\over z^2} - 2{(1+z^2)^4\over z^3}$$
+because $f'(z) = 8z(1+z^2),$ by the chain rule with $h(z) = 1+z^2 \to
+h'(z) = 2z$ and
+$k(z) = w^4 \to k'(z) = 4w^3.$
+
+\noindent{\bf 8.}
+
+\noindent{\it (a)}
+
+When $f(z) = \Re(z),$ $\delta w = \Re(z+\delta z) - \Re(z) = \Re(\delta
+z).$ $f'(z) = \delta w/\delta z = \Re(\delta z)/\delta z$ is, if $z =
+\epsilon,$ (and $\epsilon$ is real) 1, but if $z = i\epsilon,$ $f'(z) =
+0/(i\epsilon) = 0.$ Therefore, the derivative DNE.
+
+\noindent{\it (b)}
+
+Similarly, $f(z) = \Im(z)$ gives $f'(z) = \Im(\delta z)/\delta z,$
+which, for $z = i\epsilon,$ is 1, but for $z = \epsilon,$ $f'(z) = 0.$
+Therefore, the derivative also DNE.
+
+\bye
diff --git a/kang/hw4.tex b/kang/hw4.tex
new file mode 100644
index 0000000..d4fb394
--- /dev/null
+++ b/kang/hw4.tex
@@ -0,0 +1,163 @@
+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 70
+\noindent{\bf 1.}
+
+\noindent{\it (a)}
+
+If $f(z) = \overline z = x - yi,$ $U_x = 1$ and $V_y = -1.$ $U_x\neq V_y,$
+so because the Cauchy-Riemann equations are dissatisfied, the derivative
+does not exist.
+
+\noindent{\it (b)}
+
+If $f(z) = z-\overline z = x + yi - (x - yi) = 2yi,$ $U_x = 0,$ and $V_y
+= 2i.$ Again, $U_x\neq V_y.$
+
+\noindent{\it (c)}
+
+If $f(z) = 2x+ixy^2,$ 
+
+$U_x = 2.$ $V_y = 2ix.$
+$U_x = V_y \to 2 = 2ix \to x = 1/i = -i.$ $x\in R,$ so this is false,
+and the function is not analytic.
+
+\noindent{\it (d)}
+
+If $f(z) = e^xe^{-iy},$ 
+
+$$U_x = e^xe^{-iy}.$$
+$$V_y = -ie^xe^{-iy} \neq U_x.$$
+
+\noindent{\bf 3.}
+% only info from secs 21 and 23
+
+\noindent{\it (a)}
+
+$f(z) = 1/z = \overline z/|z|^2 = (a-bi)/(a^2+b^2).$
+
+The partial derivatives are $U_x = (y^2-x^2)/(x^2+y^2)^2,$
+$V_y = (y^2-x^2)/(x^2+y^2)^2,$ $U_y = (-2xy)/(x^2+y^2)^2,$ and $V_x =
+(2xy)/(x^2+y^2)^2.$ These satisfy the Cauchy-Riemann equations except
+where they are undefined, $x^2+y^2 = 0 \to z = 0.$ Using the power rule
+on $f(z) = z^{-1},$ $f'(z) = -z^{-2}.$
+
+\noindent{\it (b)}
+
+$f(z) = x^2+iy^2.$
+
+The partial derivatives are $U_x = 2x,$ $U_y = 0,$ $V_y = 2y,$ and $V_x
+= 0.$ $0 = 0$ at all points, but $2y = 2x$ only for $x = y$ or
+$z = x+ix.$
+
+Its derivative is, from these partial derivatives, $2x.$
+
+\noindent{\it (c)}
+
+$f(z) = z\Im z = (x+iy)y = xy + iy^2.$
+
+$U_x = y = V_y = 2y \to y = 0.$ $U_y = x = -V_x = 0 \to x = 0.$
+Therefore, $z = 0 + i0 = 0.$ From these partial derivatives, its
+derivative on this domain is $0.$
+
+\hrule
+%page 76
+\noindent{\bf 4.}
+
+\noindent{\it (a)}
+
+$$f(z) = {2z+1\over z(z^2+1)}$$
+
+$f(z)$ is not analytic on $z=0$ or $z^2+1=0 \to z = \pm i.$ It is,
+however, analytic on the remainder of the plane because, using the chain
+rule and the product rule, its derivative can be constructed from its
+component polynomials (which are entire functions).
+
+\noindent{\it (b)}
+
+$$f(z) = {z^3+i\over z^2-3z+2}$$
+
+This is not analytic on $z^2-3z+2 = 0 = (z-1)(z-2) \to z = 1,2.$ It is
+analytic on the remainder of the plane for the same reason as (a).
+
+\noindent{\it (c)}
+
+$$f(z) = {z^2+1\over (z+2)(z^2+2z+2)}.$$
+
+This is not analytic on $z+2 = 0 \to z = -2$ or $z^2+2z+2 = 0 =
+(z+(1-i))(z+(1+i)) \to z = 1\pm i.$ It is analytic for the same reason
+as (a).
+
+\hrule
+%page 79
+\noindent{\bf 4.}
+
+See addendum.
+
+\hrule
+%page 89
+\noindent{\bf 2.}
+
+$$f(z) = 2z^2 - 3 - ze^z + e^{-z}.$$
+
+If the component functions $f$ and $g$ are analytic on the plane (have a
+defined derivative), $f+g$ is also analytic on the plane.
+A similar rule exists for the product of entire functions.
+$2z^2$ follows the power rule, and has a derivative over the entire
+plane, and $-3$ follows the constant rule, having a derivative of 0.
+$-z$ is entire, also from the power rule, and $e^z$ has been shown to
+have a derivative in chapter 3. Using the product rule, $-ze^z$ is
+entire.
+$e^{-z}$ is a composition of $e^z$ and $-z,$ which each are analytic
+over the plane, so their composition is also analytic over the plane.
+
+This means $f(z)$ is entire.
+
+\noindent{\bf 8.}
+
+\noindent{\it (a)}
+
+$e^z = -2.$ $e^x = |z| = |-2| = 2 \to x = \ln 2,$ and
+$\arg z = \pi + 2n\pi,$ so $z = \ln2 + i(\pi+2n\pi).$
+
+\noindent{\it (b)}
+
+$e^z = 1 + i.$ $e^x = |z| = |1+i| = \sqrt 2 \to x = (1/2)\ln2,$ and
+$\arg z = \pi/4 + 2n\pi.$ Therefore, $z = (1/2)\ln2 + i(\pi/4+2n\pi).$
+
+\noindent{\it (c)}
+
+If $\exp(2z-1) = 1,$ $\exp(2z)e^{-1} = 1 \to e^{2z} = e.$ This gives
+$e^{2x} = |e| = e \to 2x = 1 \to x = 1/2,$ and $\arg(2z) = 2n\pi \to
+\arg z = n\pi,$ so $z = 1/2 + in\pi.$
+
+\hrule
+%page 95
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+$$\log e = 1 + 2n\pi i$$
+
+$\ln e = 1,$ and $\arg e = 2n\pi.$ This proves the above identity (with
+$n = 0,\pm1,\pm2,\ldots$)
+
+\noindent{\it (b)}
+
+$$\log i = \left(2n + {1\over2}\right)\pi i$$
+
+$\ln|i| = \ln1 = 0,$ and $\arg i = \pi/2 + 2n\pi.$ This makes $\log i =
+(2n+1/2)\pi i.$
+
+\noindent{\it (c)}
+
+$$\log(-1+\sqrt3i) = \ln2 + 2\left(n + {1\over3}\right)\pi i$$
+
+$\ln|-1+\sqrt3i| = \ln2,$ and $\arg(-1+\sqrt3i) = 2\pi/3 + 2n\pi,$ which
+corresponds to the given identity ($\log(-1+\sqrt3i) = (2n\pi+2\pi/3)i +
+\ln2.$)
+
+\bye
diff --git a/kang/hw5.tex b/kang/hw5.tex
new file mode 100644
index 0000000..977cbed
--- /dev/null
+++ b/kang/hw5.tex
@@ -0,0 +1,185 @@
+\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 95
+\noindent{\bf 8.}
+
+$$\log z = i\pi/2 \to z = e^{i\pi/2} = i,$$ by Euler's formula.
+
+\hrule
+%page 99
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+Where $z_n = r_ne^{i\theta_n},$ for all values of $\theta_n,$
+$$\log\left({z_1\over z_2}\right) = \ln|z_1/z_2| + i\arg(z_1/z_2)$$$$ =
+(\ln(r_1) - \ln(r_2)) + i(\theta_1-\theta_2) = (\ln(r_1) + i\theta_1) -
+(\ln(r_2) + i\theta_2) = \log z_1 - \log z_2,$$
+because $\ln(a-b) = \ln a - \ln b$ and $\arg(z_1/z_2) = \arg(z_1) -
+\arg(z_2).$
+
+\noindent{\it (b)}
+
+$1/z = \overline z/|z|^2.$ $$\log(1/z) = \ln|1/z| + i\arg(1/z) =
+- \ln|z| - i\arg z = - (\ln|z| + i\arg z) = -\log(z),$$ so
+$$\log(z*(1/z)) = \log(z) + \log(1/z) = \log(z) - \log(z) = 0.$$
+
+\hrule
+%page 103
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+The principal value of $(-i)^i$ is $e^{i\Log(i)} = e^{i\pi/2}.$
+
+\noindent{\it (b)}
+
+The base of this power function is $e^(1-2\pi/3) = b \to \Log(b) = 1-2\pi i/3.$
+This gives a principal value of $e^{3\pi i(1-2\pi/3)} = e^{3\pi i - 2\pi^2} = 
+-e^{2\pi^2}.$
+
+\noindent{\it (c)}
+
+The principal value of $(1-i)^{4i}$ is $e^{\Log(1-i)4i} =
+e^{(4i)(\ln(2)/2 - \pi/4)} = e^{(2\ln(2)i)-\pi} = e^{-\pi}(\cos(2\ln(2))
++ i\sin(2\ln(2))).$
+
+\noindent{\bf 8.}
+
+\noindent{\it (a)}
+
+$$z^{c_1}z^{c_2} = e^{c_1\Log(z)}e^{c_2\Log(z)} = e^{\Log(z)(c_1+c_2)} =
+z^{c_1+c_2}.$$
+
+\noindent{\it (b)}
+
+$${z^{c_1}\over z^{c_2}} = {e^{\Log(z)c_1}\over e^{\Log(z)c_2}} =
+e^{c_1\Log z - c_2\Log z} = e^{\Log(z)(c_1-c_2)}.$$
+
+\noindent{\it (c)}
+
+$$(z^c)^n = (e^{c\Log z})^n = e^{nc\Log z} = z^{cn}.$$
+
+\hrule
+%page 107
+\noindent{\bf 3.}
+
+$$\sin(z+z_2) = \sin z \cos z_2 + \cos z \sin z_2 \to
+\cos(z+z_2)   = \cos z \cos z_2 - \sin z \sin z_2 \to
+\cos(z_1+z_2) = \cos z_1\cos z_2-\sin z_1\sin z_2.$$
+
+\noindent{\bf 7.}
+% need to do more here
+
+$\sin z = \sin x \cosh y + i\cos x \sinh y.$
+$\cos z = \cos x \cosh y - i\sin x \sinh y.$
+
+$$|\sin z|^2 = (\sin x\cosh y)^2 + (\cos x\sinh y)^2
+= \sin^2 x(1+\sinh^2 y) + \cos^2 x\sinh^2 y =
+\sin^2 x + \sinh^2 y.$$
+
+$$|\cos z|^2 = \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y
+= \cos^2 x \cosh^2 y + (1-\cos^2 x)\sinh^2 y
+= \sinh^2 y + \cos^2 x.$$
+
+
+\noindent{\bf 12.}
+
+$\sin z$ and $\cos z$ are real with $z$ on the real axis.
+Therefore, $\overline{\sin z} = \sin\overline z$ and $\overline{\cos z}
+= \cos\overline z.$
+
+\hrule
+%page 111
+\noindent{\bf 6.}
+
+\noindent{\it (a)}
+$$|\cosh z|^2 = \sinh^2 x + \cos^2 y \to |\sinh x|^2 \leq |\cosh z|^2
+\to |\sinh x|\leq |\cosh z|.$$
+$$|\cosh z|^2 = \cosh^2 x + cos^2 y \to |\cosh z|^2 \leq \cosh^2 x \to
+|\cosh z| \leq \cosh x,$$
+because $\cosh x > 0.$
+
+\noindent{\it (b)}
+%9b of sec 38 would probably be good
+
+$$|\sinh y| \leq |\cos z| \leq \cosh y.$$
+We change $z$ to $z' = zi,$ so $y$ becomes $x.$
+$$|\sinh x| \leq |\cosh z| \leq \cosh x.$$
+
+\noindent{\bf 16.}
+$$\sinh z = \sinh x \cos y + i\cosh x \sin y.$$
+$$\cosh z = \cosh x \cos y + i\sinh x \sin y.$$
+
+\noindent{\it (a)}
+$$\sinh z = i = \sinh x\cos y + i\cosh x\sin y \to 1 = \cosh x\sin y =
+\sin y,$$
+because all zeroes of $\cosh x$ are on imaginary axis, so
+$x = 0 \to \cosh x = 1.$
+
+$\sin y = 1$ on $y = 2n\pi + \pi/2 \to z = (2n\pi + \pi/2)i.$
+
+\noindent{\it (b)}
+$$\cosh z = \cosh(yi) = 1/2 = \cos y \to y = (2n\pi \pm \pi/3)i.$$
+
+\hrule
+%page 113
+\noindent{\bf 1.}
+%learn the derivation of inverse sin
+
+\noindent{\it (c)}
+
+$$\cosh^{-1}(-1) = \log\left[-1+0^{1/2}\right] = \log(1)
+= 2n\pi i+\pi.$$
+
+\hrule
+%page 119
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+$\int_0^1 (1+it)^2 dt = \int_0^1 1 + 2it - t^2 dt =
+[t + it^2 - t^3/3]_0^1 = 1 + i - 1/3 = 2/3 + i.$
+
+\noindent{\it (b)}
+
+$\int_1^2 \left({1\over t} -i\right)^2 dt = \int_1^2 {1\over t^2} -
+{2i\over t} - 1 dt = [-{1\over t} - 2i\ln t - t]_1^2 =
+1-{1\over 2} - 2i\ln2 + 2 - 1 = 3/2 - 2i\ln2.$
+
+\noindent{\it (c)}
+
+$\int_0^{\pi/6} e^{i2t} dt = \int_0^{\pi/6} \cos(2t) + i\sin(2t) dt =
+(1/2)[\sin(2t) - i\cos(2t)]_0^{\pi/6} (1/2)[\sqrt3/2 - i/2].$
+
+\noindent{\it (d)}
+
+$\int_0^\infty e^{-zt} dt = \lim_n\to\infty [-e^{-zt}/z]_0^n = 1/z -
+\lim_t\to\infty e^{-zt}/z = 1/z.$
+
+\hrule
+%page 123
+\noindent{\bf 1.}
+%do no 2 to confirm knowledge of jordan curve
+
+\noindent{\it (a)}
+
+This is true for a real-valued function, so 
+$$\int_{-b}^{-a} u(-t) dt + i\int_{-b}^{-a} v(-t) = \int_a^b u(t) dt +
+i \int_a^b v(t) dt,$$
+because $u$ and $v$ are real-valued, so the corresponding imaginary and
+real components are equal.
+
+\noindent{\it (b)}
+
+This is true of a real function on $(a,b),$ so $$\int_a^b u(t) dt =
+\int_\alpha^\beta u(\phi(\tau))\phi'(\tau) d\tau,$$
+and similarly for $v(t),$ so
+$$\int_a^b u(t)+iv(t) dt =
+\int_\alpha^\beta (u(\phi(\tau)) + iv(\phi(\tau))\phi'(\tau) d\tau.$$
+
+\bye
diff --git a/kang/hw6.tex b/kang/hw6.tex
new file mode 100644
index 0000000..a2b684e
--- /dev/null
+++ 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
diff --git a/kang/hw7.tex b/kang/hw7.tex
new file mode 100644
index 0000000..6ca5f51
--- /dev/null
+++ b/kang/hw7.tex
@@ -0,0 +1,92 @@
+\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 147
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+The antiderivative of $z^2$ is $z^3/3$ (because the derivative of that
+function is $z^2$)
+
+$$\int_0^{1+i} z^2 dz = \left[ {z^3\over3} \right]_0^{1+i} = (2/3)(1-i).$$
+
+\noindent{\it (b)}
+
+$2\sin(z/2)$ differentiates to $\cos(z/2),$ so it is the antiderivative.
+The integral, then, evaluates to:
+
+$$2(\sin(\pi+2i) - \sin 0) = $$
+
+\noindent{\it (c)}
+
+The antiderivative of the integrand is $(z-2)^4/4,$ so the integral
+evaluates to
+$$\left[ (z-2)^4/4 \right]_1^3 = ( (3-2)^4 - (1-2)^4 )/4 = 0.$$
+
+\hrule
+%page 159
+\noindent{\bf 2.}
+
+The corollary tells us that these two integrals will be the same if the
+functions are analytic on $C_1,$ $C_2,$ and the region between them.
+
+\noindent{\it (a)}
+
+With $$f(z) = {1\over 3z^2 + 1},$$
+$f = (3z^2 + 1)^{-1} = f_1(f_2)$ can be differentiated with the chain
+rule if $f_1'$ and $f_2'$ are defined. $z^{-1} = f_1$ is analytic where
+$f_2 \neq 0,$ and $3z^2 + 1$ is a polynomial, so it is entire.
+Because $3z^2 + 1 = 0 \to |3z^2| = 1 \to |z| = \sqrt{1/3}$ is not within
+the described region (it is excluded by the square of radius 1), the
+equality holds.
+
+\noindent{\it (b)}
+
+Similarly, $$f(z) = {z+2\over \sin(z/2)}$$
+implies the equality holds if $z+2$ and $\sin(z/2)$ are analytic and
+$\sin(z/2) \neq 0.$ These are both entire functions, so the first
+criterion is trivial.
+From an earlier theorem, we know that the only zeroes of $\sin z$ are $z
+= n\pi$ with $n\in\{0,\pm 1,\pm 2,\ldots\}$ Therefore, $\sin(z/2) = 0$
+requires $z = 2n\pi.$
+However, none of these points are included in the given region because,
+where $n\neq0,$ $|z| > 4,$ so they are outside of the bounding circle,
+and when $n=0,$ $z=0$ is excluded by the square.
+
+% i would like to remember this theorem better
+% and state the proof more nicely
+
+\noindent{\it (c)}
+
+Again, $$f(z) = {z\over 1-e^z}$$
+requires $1 - e^z = 0 \to e^z = 1 \to z = \log 1 = 2n\pi i$ (with $n$
+constrained to the integers).
+These points are not contained within the circle $|z|=4,$ except $z=0,$
+which is excluded by the square of radius 1.
+
+\noindent{\bf 6.}
+
+The Cauchy-Goursat theorem doesn't apply here because this requires a
+branch cut at $\theta = -{\pi\over2},$ so $f(z) = \sqrt{r}e^{i\theta/2}$
+isn't continuously defined on the region.
+On the semicircle, $r = 1,$ $\pi \in (0, \pi),$ and $z' = ie^{i\theta},$
+so $$\int_0^\pi (\sqrt{1}e^{i\theta/2})(ie^{i\theta}) d\theta
+= \int_0^\pi ie^{3i\theta/2} d\theta
+= \left[ (2/3)e^{3i\theta/2} \right]_0^\pi
+= (2/3)(e^{3\pi i/2} - 1) = (2/3)(-i-1).$$
+
+On the two radii $\theta = 0,\pi,$ $r\in (0,1),$ the integral is
+evaluated as
+$$\int_0^1 \sqrt{r}e^{i\theta/2}(ie^{i\theta}) dr
+= e^{3i\theta/2} \int_0^1 \sqrt{r} dr
+= e^{3i\theta/2}(2/3).$$
+This is $-(2/3)i$ where $\theta = \pi,$ (evaluated in the negative
+direction, so it is $(2/3)i$ in the positive direction) and $2/3$ where
+$\theta = 0.$
+Summing these results, we get $0.$
+
+\bye
diff --git a/kang/hw8.tex b/kang/hw8.tex
new file mode 100644
index 0000000..881d5b1
--- /dev/null
+++ b/kang/hw8.tex
@@ -0,0 +1,98 @@
+\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 170
+% remember the cauchy integral proof
+\noindent{\bf 4.}
+
+Inside the contour, the extension of the Cauchy integral applies and
+tells us that $g(z)$ is equivalent to $2\pi if''(z)/2!$ where
+$f(z) = z^3 + 2z.$ $f''(z) = 6z,$ so this evaluates to $6\pi iz.$
+
+The integrand is analytic where $s\neq z,$ so with $z$ outside of $C,$
+the integral is zero by Cauchy's integral theorem---since the integrand
+is analytic inside and on the simply connected region enclosed by $C.$
+
+\noindent{\bf 6.}
+
+To determine whether $g(z)$ is analytic, we can use the definition of
+the derivative
+$$g'(z) = {g(z+\Delta z) - g(z)\over \Delta z}.$$
+If it exists, $g$ is analytic.
+$$g'(z) = {1\over 2\pi i}\int_C {f(s)\over s-z-\Delta z} - {f(s)\over
+s-z} {ds\over \Delta z}$$$$
+= {1\over 2\pi i}\int_C {f(s)ds\over (s-z-\Delta z)(s-z)}
+= {1\over 2\pi i}\int_C {f(s)ds\over (s-z)^2} + {1\over 2\pi i}\int_C
+{\Delta zf(s) ds\over (s-z-\Delta z)(s-z)^2},$$
+and that second term is limited by zero because $\Delta z$ can be
+arbitrarily low.
+
+If $f(z)$ is continuous on $C,$ where $C\cap\{z\} = \emptyset,$
+$|s-z|>0,$ so ${f(z)\over (s-z)^2}$ is continuous and the derivative
+exists.
+
+%page 177
+\noindent{\bf 2.}
+
+Because $f(z) \neq 0,$ and $f$ is analytic, $g(z) = 1/f(z)$ is
+continuous and also analytic.
+By the result for maximum values, there exists a value $M$ such that
+$|g(z)| \leq M$ for all $z\in R.$
+Because $f(z)$ is not constant, $g(z)$ also isn't constant, and the only
+points such that $g(z_0) = M$ are on the boundary.
+
+These conditions imply that, for all $z\in R,$ $|f(z)| \geq 1/M,$ and
+the only points such that $f(z_0) = 1/M$ is on the boundary.
+This proves that the minimum of $f$ are on the boundary.
+
+\noindent{\bf 4.}
+
+$$|f(z)|^2 = |\sin z|^2 = \sin^2 x + \sinh^2 y$$
+is maximized when $\sin x$ is maximized and when $\sinh y$ is maximized.
+$\sin x$ has a maximum at $x=\pi/2$ on $(0,\pi)$ because this is the
+maximum of the real function.
+$\sinh y$ has a maximum at $y=1$ on $(0,1)$ because $\sinh x$ tends to
+infinity as $x$ approaches either positive or negative infinity.
+Therefore, $|f(z)|$ is maximized at $x + yi = \pi/2 + i.$
+
+%page 185
+\noindent{\bf 2.}
+
+The principal arguments of these numbers are, for $\Theta_{2n},$
+$\tan^{-1}(1/n^2),$ and for $\Theta_{2n+1},$ $\tan^{-1}(-1/n^2).$
+Both of these are arbitrarily close to 0 for sufficiently large $n,$ so
+the series has sum $0.$
+
+This series converges unlike Example 2 in Section 60 because the value
+to which it converges is not a branch cut, where a given point can be
+infinitely close to distinct values of $\Theta.$
+
+\noindent{\bf 3.}
+
+If $$\lim_{n\to\infty} z_n = z,$$
+
+$|z_{n} - z| < \epsilon$ for $n > n_0.$
+Because $||z_n| - |z|| \leq |z_n - z|,$ $||z_n| - |z|| < \epsilon$ also
+for $n > n_0.$
+
+This shows that
+$$\lim_{n\to\infty} |z_n| = |z|.$$
+
+\noindent{\bf 8.}
+
+$$\sum_{n=1}^\infty z_n = S = S_x + iS_y,$$
+which are in turn defined as infinite sums over the real numbers.
+Similarly, $T = T_x + iT_y.$
+
+Therefore,
+$$\sum_{n=1}^\infty (z_n + w_n) = \sum_{n=1}^\infty (z_{x_n} + iz_{y_n}
++ w_{x_n} + iw_{y_n}) = \sum_{n=1}^\infty (z_{x_n} + w_{x_n}) +
+i(z_{y_n} + w_{y_n})$$$$ = (S_x + T_x) + i(S_y + T_y) = S_x + iS_y + T_x
++ iT_y = S + T,$$
+made possible by the analogous property for real series and the fact
+that $\sum_{n=1}^\infty iz_n = i\sum_{n=1}^\infty z_n.$
+
+\bye
diff --git a/kang/hw9.tex b/kang/hw9.tex
new file mode 100644
index 0000000..10ba2f7
--- /dev/null
+++ b/kang/hw9.tex
@@ -0,0 +1,185 @@
+\def\Log{\mathop{\rm Log}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 195
+\noindent{\bf 1.}
+
+First,
+$$\cosh z = \sum_{n=0}^\infty {z^{2n}\over (2n)!}$$
+
+Substituting $z^2$ for $z$ and then multiplying each term by $z$ will
+give $z\cosh(z^2).$
+
+$$z{(z^2)^{2n}\over (2n)!} = {zz^{4n} \over (2n)!} = {z^{4n+1}\over
+(2n)!}.$$
+
+This gives
+$$z\cosh(z^2) = \sum_{n=0}^\infty {z^{4n+1}\over (2n)!}$$
+This applies over the entire plane because this function is entire.
+
+\noindent{\bf 3.}
+
+The Maclaurin expansion of
+$$f(z) = {z\over z^4+4} = {z\over 4}\cdot {1\over 1+(z^4/4)}$$
+may be found by substituting $-z^4/4$ in the place of $z$ in the
+standard $1/(1-z)$ Maclaurin expansion, then multiplying by $z/4.$
+
+This gives
+$$f(z) = \sum_{n=0}^\infty {z^{4n+1}\over 4^{n-1}}$$
+
+\noindent{\bf 7.}
+
+$$f(z) = \sin z = {e^{iz} - e^{-iz}\over 2}.$$
+$$f^{(n)}(z) = i^n{e^{iz} - (-1)^ne^{-iz}\over 2}.$$
+Where $n = 2k+1,$ where $k$ is an integer, $f^{(n)}(0)$ simplifies to
+$0$ because $e^{i0} - (-1)^{2k+1}e^{-i0} = 1 - 1 = 0.$
+Similarly, where $n = 2k,$ $f^{(n)}(0)$ simplifies to $i^{2k} = (-1)^k$
+because $e^{i0} - (-1)^{2k}e^{-i0} = 1 + 1 = 2,$ which substituted into
+the original equation becomes $i^{2k}{2\over 2}.$
+
+This provides an alternate justification of the Maclaurin expansion
+given in Section 64.
+Using the standard Maclaurin sum,
+$$f(z) = \sum_{n=0}^\infty {f^{(n)}(0)z^n\over n!} = \sum_{k=0}^\infty
+{f^{(2k)}(0)z^{2k}\over (2k)!} + \sum_{k=0}^\infty
+{f^{(2k+1)}(0)z^{2k+1}\over (2k+1)!} = \sum_{k=0}^\infty
+{(-1)^kz^{2k}\over (2k)!}$$
+
+\hrule
+%page 205
+\noindent{\bf 2.}
+
+$$f(z) = {1\over 1+z} = {1\over z}\cdot {1\over 1+(1/z)}$$
+
+By substitution into the standard $1/(1-z)$ expansion (with $-1/z$ being
+substituted for $z$),
+$${1\over 1+(1/z)} = \sum_{n=0}^\infty (-1)^n(1/z)^n =
+\sum_{n=0}^\infty {(-1)^n\over z^n}.$$
+$${1\over z} \cdot {1\over 1+(1/z)} = \sum_{n=0}^\infty {(-1)^n\over
+z^{n+1}} = \sum_{n=1}^\infty {(-1)^{n-1}\over z^n}$$
+
+The standard $1/(1-(-1/z))$ applies because
+$1<|z|<\infty \to |-1/z| < 1.$
+
+\noindent{\bf 3.}
+
+$${1\over z(1+z^2)} = {1\over z^3(1+(1/z^2))}$$
+
+With $1<|z|<\infty \to |-1/z^2|<1,$
+$${1\over (1-(-1/z^2))} = \sum_{n=0}^\infty (-1)^n z^{-2n}.$$
+$${1\over z^3(1-(-1/z^2))} = \sum_{n=0}^\infty (-1)^n z^{-2n-3}.$$
+
+\noindent{\bf 5.}
+
+% I know how to do this one. Just formulate 1/(z-1) and repeat for
+% 1/(z-2)
+
+$$f(z) = {-1\over (z-1)(z-2)} = {1\over z-1} - {1\over z-2}.$$
+
+Within $|z|<1,$
+$${1\over z-1} = -{1\over 1-z} = \sum_{n=0}^\infty (-1)z^n.$$
+On $|z|>1,$
+$${1\over z-1} = {1\over z}\cdot{1\over 1 - (1/z)} =
+\sum_{n=0}^\infty z^{-(n+1)}.$$
+
+Within $|z|<2,$
+$${1\over z-2} = -{1\over 2}\cdot{1\over 1 - z/2} =
+\sum_{n=0}^\infty {(-1)z^n\over 2^{n+1}}.$$
+On $|z|>2,$
+$${1\over z-2} = {1\over z}\cdot{1\over 1-(2/z)} = \sum_{n=0}^\infty
+2^n/z^(n+1)$$
+
+Within $|z|<1$ and $|z|<2,$ ($D_1$) these become
+$$\sum_{n=0}^\infty z^n\left({1\over 2^{n+1}} - 1\right).$$
+
+Within $|z|>1$ and $|z|<2,$ ($D_2$) these become
+$$\sum_{n=0}^\infty {z^n\over 2^{n+1}} + \sum_{n=1}^\infty {1\over z^n}.$$
+
+Within $|z|>1$ and $|z|>2,$ ($D_3$) these become
+$$\sum_{n=1}^\infty {1 - 2^{n-1}\over z^n}$$
+
+\hrule
+%page 218
+\noindent{\bf 1.}
+
+The first derivative of $1/(1-z)$ is $1/(1-z)^2,$ which will have the
+Maclaurin series equal to the termwise differentiation of the $1/(1-z)$
+Maclaurina series. Except $n = 0,$ (for which $z^n = z^0$ has first
+derivative 0) the first derivative of $z^n$ is $nz^{n-1}.$
+This gives summation
+$${1\over (1-z)^2} = \sum_{n=1}^\infty nz^{n-1} = \sum_{n=0}^\infty
+(n+1)z^n.$$
+
+Similarly, the second derivative of $1/(1-z)$ is $2/(1-z)^3.$
+The second derivative of $z^n$ for $n\geq2$ is $n(n-1)z^{n-2},$ giving
+summation
+$${2\over (1-z)^3} = \sum_{n=2}^\infty n(n-1)z^{n-2} = \sum_{n=0}^\infty
+(n+2)(n+1)z^n.$$
+
+\noindent{\bf 5.}
+
+$${\cos z\over z^2 - (\pi/2)^2} = {\cos z\over (z-\pi/2)(z+\pi/2)} =
+{\cos z\over \pi(z-\pi/2)} - {\cos z\over \pi(z+\pi/2)} =
+{-\sin(z-\pi/2)\over \pi(z-\pi/2)} - {\sin(z+\pi/2)\over \pi(z+\pi/2)}.$$
+
+As shown in the example in Section 71, $\sin w/w$ is analytic, so this
+function is analytic.
+
+\noindent{\bf 6.}
+
+With $|z-1|<1,$
+$$\int_1^z {1\over w}dw = \sum_{n=0}^\infty \int_1^z (-1)^n (w-1)^n dw
+\to \Log z = \sum_{n=0}^\infty (-1)^n (z-1)^{n+1}/(n+1) =
+\sum_{n=1}^\infty {(-1)^{n+1}(z-1)^n\over n}.$$
+
+\hrule
+%page 224
+\noindent{\bf 1.}
+
+On $0<|z|<1,$
+$$e^z = 1 + z + z^2/2 + z^3/6 + \cdots$$
+
+$${1\over z(z^2+1)} = 1/z - z + \cdots$$
+
+Taking their product gives
+
+$${e^z\over z(z^2+1)} = 1/z + 1 + z/2 + z^2/6 - z - z^2 + \cdots =
+1/z + 1 - z/2 - 5z^2/6 + \cdots$$
+
+\hrule
+%page 237
+\noindent{\bf 1.}
+% I don't understand residues at all.
+
+\noindent{\it (a)}
+
+$${1\over z+z^2} = {(1+z)^{-1}\over z},$$
+giving a residue of $1+z,$ or, at $z=0,$ $1.$
+
+\noindent{\it (b)}
+
+$$z\cos\left({1\over z}\right) = \sum_{n=0}^\infty (-1)^n {z^{1-2n}\over
+(2n)!}.$$
+The value at $z^{-1}$ or $1-2n = -1 \to n = 1$ is $-1/2.$
+
+\noindent{\it (c)}
+
+$${z-\sin z\over z} = 1 - \sum_{n=0}^\infty (-1)^n {z^{2n}\over (2n+1)!}$$
+has residue of $0$ because this is strictly a Taylor series.
+
+\noindent{\it (d)}
+
+Noting that $\cot z = 1/z - z/3 - z^3/45 + \ldots,$
+$${\cot z\over z^4} = {1\over z^4}(1/z - z/3 - z^3/45 + \ldots) =
+{1\over z^5} - {1\over 3z^3} - {1\over 45z} + \ldots,$$
+giving a residue of $-1/45.$
+
+\noindent{\it (e)}
+
+$${\sinh z\over z^4(1-z^2)} =
+z^{-4}(z+{z^3\over 3!}+\cdots)(1+z^2+z^4+\cdots) =
+z^{-3} + z^{-1}/6 + z^{-1} + z/6 + z + z^3/6,$$
+giving a residue of $7/6.$
+
+\bye
-- 
cgit