\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