1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
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
|