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+\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