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%\markright{Math 428 Exam I, Spring 2004\hfill Page }

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\begin{document}
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\noindent {\bf \rule{\textwidth}{.01in}
\newline
\hfill Solution Key to Selected Exercises of MATH 423/823 \hfill
\newline
\rule[.5em]{\textwidth}{.01in}
\newline}
%\bigskip

\smallskip
%\newline
%\rule{\textwidth}{.01in}

\bigskip\noindent\textbf{Sec.1.2.}

14. Prove that $|z_1z_2|=|z_1||z_2|$.

\noindent{\it Proof:} Since $z^2=z\overline{z}$ and
$\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}$, then by the
commutativity and associativity of multiplication we have
$|z_1z_2|^2=({z_1z_2})(\overline{z_1z_2})=({z_1\overline{z_1}})(z_2\overline{z_2})=|z_1|^2|z_2|^2$,
which implies the identity.

\bigskip\noindent\textbf{Sec.1.3.}

16. Prove that $||z_1|-|z_2||\le|z_1-z_2|$.

\noindent{\it Proof:} Let $z=z_1-z_2$ and $w=z_2$. Since $z+w=z_1$,
we have by the triangle inequality that $|z_1|=|z+w|\le
|z|+|w|=|z_1-z_2|+|z_2|$. Hence, $|z_1|-|z_2|\le|z_1-z_2|$.
Exchanging the roles of $z_1$ and $z_2$, we have
$-(|z_1|-|z_2|)=|z_2|-|z_1|\le|z_2-z_1|=|-(z_1-z_2)|=|z_1-z_2|$,
equivalently, $-|z_1-z_2|\le |z_1|-|z_2|$. Together, the
inequalities  $-|z_1-z_2|\le |z_1|-|z_2|$ and
$|z_1|-|z_2|\le|z_1-z_2|$ is equivalent to $||z_1|-|z_2||\le
|z_1-z_2|$. The end of proof.

\bigskip\noindent\textbf{Sec.1.6.}

10. Prove that the closed disk $|z-z_0|\le \rho$ is bounded.

\noindent{\it Proof:} For every $z$ satisfying $|z-z_0|\le \rho$ we
have $||z|-|z_0||\le|z-z_0|\le \rho$ which is $-\rho\le |z|-|z_0|\le
\rho$. The second inequality becomes $|z|\le |z_0|+\rho:=N$, showing
that $z$ is bounded by $N$.

13. Let $S$ be a subset of $\mathbb C$. Prove that $S$ is closed if
and only if its complement $S^c:={\mathbb C}\setminus S$ is open.

\noindent{\it Proof:} Denote by $\partial D$ the boundary of a set
$D$. It is straightforward to show by definition that $\partial
S=\partial S^c$ because $\forall z\in \partial S$ and $\forall
\epsilon>0,\ D_\epsilon(z)\cap S\ne \emptyset$ and
$D_\epsilon(z)\cap S^c\ne \emptyset$ which by definition is
equivalent to $z\in S^c$. To show $S^c$ is closed if $S$ is open we
only need to show $\forall z\in\partial S^c \Rightarrow z\in S^c$.
This is true because if $z\in S$ then $\exists D_\rho(z)\subset S$
for some $\rho>0$ since $S$ is open, an contradiction. To show $S$
is open if $S^c$ is closed we only need to show for each $z\in S$
there is a $\rho$ so that $D_\rho(z)\subset S$. This is true because
$z\centernot\in S^c \Rightarrow z\centernot\in \partial S^c=\partial
S \Rightarrow z\centernot\in \partial S \Rightarrow $ there must be
one $\rho>0$ so that $D_\rho(z)\subset S$. Otherwise, we would have
for all small $\epsilon>0$ there must be at least one point in
$D_\epsilon(z)\cap S^c$, a contradiction.

\bigskip\noindent\textbf{Sec.1.7.}

8. Second part: give a geometric explanation of
$\chi(z,w)=\chi(1/z,1/w)$.

\noindent{\it Outline of Explanation:} Let $R_2$ be the reflection
operation with respect to the $x_1x_3$-plane in $\mathbb R^3$,
$R_2(Z)=R_2(x_1,x_2,x_3)=(x_1,-x_2,x_3)$ and similarly $R_3$ be the
reflection operation with respect to the $x_1x_2$-plane with
$R_3(Z)=R_3(x_1,x_2,x_3)=(x_1,x_2,-x_3)$. Let $Z=P(z)$ denote the
stereographic projection. We note first that both $R_2$ and $R_3$
preserve the Euclidean distance: $d(Z,W)=d(R_i(Z),R_i(W))$. We note
second that the $x_1x_3$-reflection reserves the stereographic
projection, i.e. $R_2(P(z))=P(R_2(z))$, because $R_2$ transforms
$z,Z=P(z)$ and its defining line through the north pole to the
corresponding reflections. We should also note that the two
reflection operations map points from the unit sphere to points on
the unit sphere. Let $Z=P(z),Z'=R_2(Z),Z''=R_3(Z')$ and similar
notations for $w$. We know $d(Z,W)=d(Z',W')=d(Z'',W'')$. It only
remains to show this claim that $P(\frac{1}{z})=Z''=R_3(R_2(P(z)))$,
which implies $\chi (z,w)=d(Z,W)=d(Z'',W'')=\chi(1/z,1/w)$. Use the
polar coordinate form $z=re^{i\theta}$ to get
$1/z=\frac{1}{r}e^{-\theta}$, a point on the radial plane with the
phase angle $-\theta$ in $\mathbb R^3$. The reflected point
$z'=R_2(z)$ and $Z'=R_2(Z)$ lie on the same radial plane as well.
The claim is proved by using Euclidean geometry as depicted in the
diagram below. It is a cross-section view on the radial plan ${\rm
Arg} z=-\theta$ for the case of $r\ge 1$. From which we see the
following: $\Delta NOD\sim \Delta NSZ''$ because they are right
triangles and share a common angle. Also $\Delta NSZ''\sim \Delta
NOr$ because they are right triangles and because the angle $\angle
NSZ''=\angle SNr$ as they are from the isosceles triangle $\Delta
NSr$. Therefore $\Delta NOD\sim \Delta NOr$ which implies $D/1=1/r$
by similarity. Thus $P(1/z)=Z''$. For the case of $r<1$, the same
diagram works after exchanging $r$ with $1/r$ and $Z'$ with $Z''$.
This completes the outline.

\begin{tikzpicture}[scale=1.5]
\node (O) at (0, 0) {};

\node (D) at ({(1-(sqrt(2)/2))/(sqrt(2)/2)}, 0) {};

\node (N) at (0,1) {};

\path [fill=blue!10] (0,0) to ({(1-(sqrt(2)/2))/(sqrt(2)/2)},0) to
(0,1);

\draw[->] (-1.2,0) -- (3,0) node [below] {$\mathbb C$};

\draw[->] (0,-1.2) -- (0,1.2) node [above] {$x_3$};

\draw[color=gray,line width=.1mm, dashed] (O) circle (1);

\node at (-.2, 1) [above] {$N$};

\node at (-0, 1/2) [left] {$1$};

\node at (-.2, 0) [below] {$O$};

\node at (-.2, -1) [below] {$S$};


\def\r{.6};
\def\u{1/sqrt(2)};
\def\v{1/sqrt(2)};
%\def\t{sqrt(1-\u^2)};
%\def\z{\sqrt{1-\x*\x}};
%\def\z{.5};

%\node at (.14, .14) [above] {$Z'$};

\node at ({sqrt(2)/2}, {sqrt(2)/2}) [above] {$Z'$};

\node at ({sqrt(2)/2}, {-sqrt(2)/2}) [below] {$Z''$};

\node at ({(sqrt(2)/2)/(1-(sqrt(2)/2))-.1}, .2) [right]
{$r=|R_2(z)|$};

\node at ({(1-(sqrt(2)/2))/(sqrt(2)/2)+.1}, 0) [above] {$D$};

\node at (.2, 0) [below] {$\frac{1}{r}$};

\draw[color=red,thick] (0, 1) -- ({(sqrt(2)/2)/(1-(sqrt(2)/2))}, 0);

\draw[color=red,thick] (0, 1) -- ({sqrt(2)/2}, {-sqrt(2)/2});

\draw[color=blue,thick] ({sqrt(2)/2}, {sqrt(2)/2}) -- ({sqrt(2)/2},
{-sqrt(2)/2});

\draw[color=gray,dashed] (0, -1) -- ({(sqrt(2)/2)/(1-(sqrt(2)/2))},
0);


\draw [fill] (O) circle [radius=.05];

\draw[color=gray,line width=.1mm, dashed] (O) circle (1);

\end{tikzpicture}

\bigskip\noindent\textbf{Sec.2.2.}

Theorem 1(ii): If $\lim_{z\to z_0}f(z)=A,\ \lim_{z\to z_0}g(z)=B$,
then $\lim_{z\to z_0}f(z)g(z)=AB$.

\noindent{\it Proof:} We first show that If $\lim_{z\to z_0}f(z)=A$
then $f$ is bounded by a constant $M$ in a disk $D_\rho(z_0)$ for
some $\rho>0$. In fact, by definition for $\epsilon =1$ there is a
$\rho>0$ so that $|z-z_0|<\rho\Rightarrow |f(z)-A|<\epsilon=1$ which
implies $||f(z)|-|A||\le|f(z)-A|<1$ and $|f(z)|<|A|+1:=M$ follows.
Thus, without loss of generality (wlog), we can assume
$|f(z)|,|g(z)|\le N$ for $z\in D_r(z_0)$ for some number
$N>\max{|A|+1,|B|+1}$ and $r>0$.

We first use the assumption to have $\forall \epsilon>0\ \exists\
\delta_1>0 {\rm \ and \ } \delta_2>0$ with $\delta_{1,2}<r {\rm\
s.t. (such\ that)\ } |z-z_0|<\delta_1 \Rightarrow
|f(z)-A|<\epsilon/(2N) {\rm \ and \ } |z-z_0|<\delta_2 \Rightarrow
|g(z)-B|<\epsilon/(2N)$. Now let $\delta=\min\{ \delta_1,\delta_2\}$
then $\forall |z-z_0|<\delta\le \delta_1,\delta_2$ both
$|f(z)-A|<\epsilon/(2N)$ and $|g(z)-B|<\epsilon/(2N)$.

Now for the product rule, we have $\forall z {\rm \  with\ }
|z-z_0|<\delta$
\[
|f(z)g(z)-AB|=|f(z)g(z)-Ag(z)+Ag(z)-AB|\le
|f(z)-A||g(z)|+|A||g(z)-B|
\]
\[
\qquad \le
N(|f(z)-A|+|g(z)-B|)<N[\epsilon/(2N)+\epsilon/(2N)]=\epsilon.
\]
This proves the result.

22. Show that if $\lim_{n\to\infty}f(z_n)=w_0$ for every sequence
$\{z_n\}_1^\infty$ converging to $z_0$ ($z_n\ne z_0$), then
$\lim_{z\to z_0}f(x)=w_0$.

\noindent{\it Proof:} Suppose on the contrary that $\lim_{z\to
z_0}f(x)\ne w_0$. Then by definition there must be a number
$\epsilon_0>$ so that for every $\delta>0$ the inequality
$|f(z)-w_0|<\epsilon_0$ does not hold for every point
$|z-z_0|<\delta$. That is there is some point $z_\delta$ with
$|z_\delta-z_0|<\delta$ but $|f(z_\delta)-w_0|\ge \epsilon_0$. In
particular, for each integer $n$ we can find a point $z_n$
satisfying $|z_n-z_0|<1/n$ but $|f(z_n)-w_0|\ge \epsilon_0$. This
contradicts the assumption because by construction this sequence
$z_n\to z_0$ but $\lim_{n\to\infty}f(z_n)\ne w_0$.

\bigskip\noindent\textbf{Sec.2.4.}

Prove the product rule of derivative:
$[f(z)g(z)]'=f'(z)g(z)+f(z)g'(z)$

\noindent{\it Proof:} We will use the fact that a function is
differentiable at a point is also continuous at the point. By
definition, and a standard trick of adding and subtracting a same
term, $f(z)g(z+\Delta z)$, in the numerator below, we have
\begin{equation*}
\begin{split}
[f(z)g(z)]'&=\lim_{\Delta z\to 0}\frac{f(z+\Delta z)g(z+\Delta
z)-f(z)g(z)}{\Delta z}\\
&= \lim_{\Delta z\to 0}\frac{f(z+\Delta z)g(z+\Delta
z)-f(z)g(z+\Delta z)+f(z)g(z+\Delta z)-f(z)g(z)}{\Delta z}\\
&= \lim_{\Delta z\to 0}\left[\frac{f(z+\Delta z)-f(z)}{\Delta z}
g(z+\Delta z)+f(z)\frac{g(z+\Delta z)-g(z)}{\Delta z}\right]\\
&= \lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}
\lim_{\Delta z\to 0}g(z+\Delta z)+f(z)\lim_{\Delta z\to 0}\frac{g(z+\Delta z)-g(z)}{\Delta z}\\
&=f'(z)g(z)+f(z)g'(z).
\end{split}
\end{equation*}
The last equality follows from the assumption and the fact that all
the limits involved exist, as well as the production and summation
rules of limit. This proves the result.

11 Suppose that $f(z)$ and $\overline{f(z)}$ are analytic in a
domain. Show that $f(z)$ is constant in $D$.

\noindent{\it Proof:} Let $f(z)=u(x,y)+iv(x,y)$ and
$g(z)=f(z)+\overline{f(z)}=2u(x,y)$. Then by the summation rule, $g$
is also analytic in $D$. Since ${\rm Im}g(z)=0$, by the C-R
equations we have $u_x=u_y=0$ and hence $u(x,y)$ is constant, which
in turn implies $v(x,y)$ is constant, and hence the proof.

\bigskip\noindent\textbf{Sec.3.3.}

Lecture example: Find the domain of the function $w=z^2$ whose range
is the plane with the branch cut $L=\{{\rm Re}w=1,{\rm Im}w\in
[0,+\infty)\}$.

\noindent{\it Solution:} The branch cut $L$ has two sides. On the
left side $w\in L_-$, we have ${\rm arg}w={\rm Arg}w\in [0,\pi/2)$.
On the right side $w\in L_+$, we have ${\rm arg}w={\rm Arg}w+2\pi\in
[2\pi,2\pi+\pi/2)$. To find the preimage of $L_{\pm}$ is to find the
image of the inverse map $z=w^{1/2}=|w|^{1/2}e^{i({\rm
arg}w)/2}=|w|^{1/2}e^{i({\rm Arg}w/2+k\pi)}$. Let $w=u+iv$ and ${\rm
Arg}w=\theta$. Then for $w\in L_-$ with $k=0$, $|w|=\sqrt{1+v^2}$,
and $\cos(\theta/2)=\sqrt{\frac{1+\cos(\theta)}{2}}$,
$\sin(\theta/2)=\sqrt{\frac{1-\cos(\theta)}{2}}$ by the half-angle
formula. As $\tan\theta=v/u=v$, we have
$\cos(\theta)=1/\sqrt{1+v^2}$ (and $\sin(\theta)=v/\sqrt{1+v^2}$).
Therefore
$x+iy=z=w^{1/2}=(1+v^2)^{1/4}[\cos(\theta/2)+i\sin(\theta/2)]=(1+v^2)^{1/4}[
\sqrt{\frac{\sqrt{1+v^2}+1}{2\sqrt{1+v^2}}}
+i\sqrt{\frac{\sqrt{1+v^2}-1}{2\sqrt{1+v^2}}}]$, which is simplified
to $x=s(v)=\sqrt{\frac{\sqrt{1+v^2}+1}{2}}$ and
$y=t(v)=\sqrt{\frac{\sqrt{1+v^2}-1}{2}}$ parameterized by parameter
$v\in [0,\infty)$. This obtains $L_-$ corresponding side of a cut in
the domain $D$. For the $L_+$ side of the cut in the range, since
${\rm arg}w={\rm Arg}w+2\pi$, the image under $w^{1/2}$ differs from
that of $L_-$ exactly by a negative sign, that is $x=-s(v),y=-t(v)$.
See figure, in which the additional cut $\{0\le u\le 1,\ v=0\}$ is
needed for the definition of the inverse function $z=w^{1/2}$. Also
the corresponding subregions in domain and range are color coded.

\begin{tikzpicture}[scale=.75]

\def\xrange{{sqrt((sqrt(1+18^2)+1)/2)}};

\def\yrange{{sqrt((sqrt(1+18^2)-1)/2)}};

\path [fill=blue!10, domain=0:18] plot ({sqrt((sqrt(1+\x^2)+1)/2)},
{sqrt((sqrt(1+\x^2)-1)/2)}) to (0,\yrange) to (0,0) to (1,0);

\path [fill=gray!20, domain=0:18] (0,0) to (0, \yrange) to
(-\xrange,\yrange) to (-\xrange,0);

\path [fill=red!10, domain=0:18] plot
({-sqrt((sqrt(1+(18-\x)^2)+1)/2)}, {-sqrt((sqrt(1+(18-\x)^2)-1)/2)})
to (-\xrange,0);

\path [fill=gray!20] (10.8, 0) to (5.2,0) to (5.2,-2.8) to
(10.8,-2.8);

\path [fill=blue!10] (5.2,0) to (9,0) to (9,3) to (5.2,3);

\path [fill=red!10] (10.8, 3) to (9,3) to (9,0) to (10.8,0);

\draw[color=white,ultra thick] (8,0) to (9,0) to (9,3);

\node (O) at (0, 0) {};

\draw[-] (-3.3,0) -- (-1,0);

\draw[->] (1,0) -- (3,0) node [below] {$x$};

\draw[->] (0,-3) -- (0,3) node [above] {$y$};

\draw[->] (5,0) -- (11,0) node [below] {$u$};

\draw[->] (8,-3) -- (8,3.1) node [above] {$v$};

\draw[->,red] (9.05,3) -- (9.05,-.05);

\draw (9.05,-.05) node [below] {$1$};

\draw[->,blue] (8.95,.1) -- (8.95,3);

\draw[->,red,dashed] (8.95,-.05) -- (8,-.05);

\draw[->,blue,dashed] (8,.05) -- (8.95,.05);

\draw[-,dashed] (-3, -3) -- (3,3);

\draw [->, blue, domain=0:14] plot ({sqrt((sqrt(1+\x^2)+1)/2)},
{sqrt((sqrt(1+\x^2)-1)/2)});

\draw [->, red, domain=14:0] plot ({-sqrt((sqrt(1+\x^2)+1)/2)},
{-sqrt((sqrt(1+\x^2)-1)/2)});

\draw[->,red,dashed] (-1,-.0) -- (0,-.0);

\draw[->,blue,dashed] (0,-.0) -- (1,-.0);

\draw (1,-.0) node[below] {$1$};

\draw (-1,-.0) node[above] {$-1$};

\draw[->] (3.5,1) -- (4.5,1);

\draw (4,1.0) node[above] {$w=z^2$};

\draw[<-] (3.5,-1) -- (4.5,-1);

\draw (4,-1.0) node[below] {$z=w^{1/2}$};

\draw (7.8,-.0) node[below] {$0$};

\draw (.2,-.0) node[below] {$0$};

\draw[blue] (9,2) node[left] {$L_-$};

\draw[red] (9,2) node[right] {$L_+$};

\end{tikzpicture}

\bigskip\noindent\textbf{Sec.3.3.}

10. Show that the function ${\rm Log}(-z)+i\pi$ is a branch of $\log
z$ in the domain $D_0$ consisting of all points in the plane except
those on the nonnegative real axis.

\noindent{\it Solution:} By definition, ${\rm Log}(-z)+i\pi=\log|z|
+i({\rm Arg}(-z)+\pi)$. There are two cases as ${\rm Arg}(z)\in
(-\pi,\pi]$. For $z$ in the quadrant I or II, $-z$ is in quadrant
III or IV, ${\rm Arg}(-z)={\rm Arg}(z)-\pi$. For $z$ in the quadrant
III or IV, $-z$ is in quadrant I or II, ${\rm Arg}(-z)={\rm
Arg}(z)+\pi$. Therefore, ${\rm Arg}(-z)+\pi = {\rm Arg}(z)={\rm
arg}_0(z)\in (0,\pi)$ if $z$ is in quadrant I or II, and ${\rm
Arg}(-z)+\pi = {\rm Arg}(z)+2\pi ={\rm arg}_ 0(z)\in (\pi,2\pi)$ if
$z$ in the quadrant III or IV. Together, this shows ${\rm
Log}(-z)+i\pi=\log _0(z)$.


\end{document}
