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
|
\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
|
