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\begin{document}

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\begin{center}
{\bf Conformal Mappings and the Schwarz-Christoffel Transformation}\vspace{20pt}

\rule{4in}{1pt}\vspace{20pt}

{\bf A Thesis

 presented to

 the Faculty of the Graduate School

 University of Missouri}\vspace{20pt}

\rule{4in}{1pt}\vspace{20pt}

In Partial Fulfillment

of the Requirements for the Degree

Master of Arts \vspace{20pt}

\rule{4in}{1pt}\vspace{20pt}

by

Rebecca Heinen

Dr. Loukas Grafakos, Thesis Supervisor \vspace{20pt}

May 2017\end{center}\newpage


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The undersigned, appointed by the Dean of the Graduate School, have examined
the thesis entitled

\begin{center} Conformal Mappings and the Schwarz-Christoffel Transformation \end{center}

presented by Rebecca Heinen, a candidate for the degree of Master of Arts, and hereby certify that in their opinion it is worthy of acceptance.

\hspace{1.875in}\parbox{4in}{\begin{center}

\vspace{42pt}\rule{4in}{1pt}\vspace{.125in}\\Professor Loukas Grafakos

\vspace{42pt}\rule{4in}{1pt}\vspace{.125in}\\Professor Nakhl\' e Asmar

\vspace{42pt}\rule{4in}{1pt}\\Professor David Retzloff \end{center}}
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\begin{center}ACKNOWLEDGEMENTS \end{center}

\indent I would like to thank Dr. Loukas Grafakos for allowing me to work on this project with him and providing me with invaluable academic and career advisement.  This thesis is largely based on the results provided in Chapter 6 of \textit{Applied Complex Analysis with Partial Differential Equations} by Dr. Nakhl\' e Asmar and the second edition by Dr. Asmar and Dr. Grafakos \cite{AsmarGrafakos}.  To understand the material on a deeper level, I wrote solutions for a future solution manual for the exercises presented in the sections covered by this thesis.  A select number of these exercises were chosen to serve as propositions, lemmas, or examples.  Undertaking this project would not have been possible without Dr. Asmar's book, so I would like to also thank him for allowing me to work closely with the text and allowing me to use the figures from that chapter.  I also want to thank Dr. Asmar and Dr. David Retzloff for taking the time to serve on my thesis committee.  \\
\indent I am so incredibly thankful for all the friends, family, and previous educators who supported me along the way.  I'm greatful for the colleagues/friends I have made in the past two years.  We have struggled emotionally and academically, but I'm glad we banded together to overcome these struggles and become better mathematicians!  I am thankful that North Central College provided me with a strong network of faculty, friends, and mentors that encouraged me to attend and finish graduate school.  The faculty members at NCC serve as great role models for the educator I aspire to be.  Lastly, I would like to thank my family, especially my grandmother, for supporting me emotionally through all the ups and downs of graduate school.  
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\begin{center} Conformal Mappings and the Schwarz-Christoffel Transformation

Rebecca Heinen

Dr. Loukas Grafakos, Thesis Supervisor 

ABSTRACT
\end{center} 

\indent Let $\Omega$ be an open and connected set in the complex plane.  A mapping $f:\Omega \rightarrow \mathbb{C}$ is said to be conformal at a point $z_{0}$ if it preserves angles and orientation between curves intersecting at $z_{0}$.  We discuss tangent lines and their relationship to the conformal property.  This property is useful for determining the points at which a function is conformal.  We provide several examples and a summary of boundary behavior properties.  \\
\indent We discuss two types of conformal mappings, linear fractional transformations and the Schwarz-Christoffel transformation.  Linear fractional transformations map lines and circles to either lines or circles; meanwhile, the Schwarz-Christoffel transformation maps the upper half-plane onto a polygon $P$.  We prove several properties of these mappings and demonstrate their importance with the use of examples.  \\
\indent These results are useful for solving boundary value problems on irregular regions.  If we map such a region onto another region whose solution is known, we are able to then obtain a solution on the first region.  Important physical applications of this thesis include determination of isotherms, streamlines, and air gap reluctance for magnetic circuits.  
\addcontentsline{toc}{chapter}{Abstract}
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\chapter{Introduction to Conformal Mappings}  \lb{ch1}
Our goal in this chapter is to introduce basic properties of mappings given by analytic functions.  We will begin with a brief recall of tangent lines. 

\section*{Basic Properties} 

Let $\gamma$ be a smooth path parametrized by $z(t)=x(t)+iy(t)$ with $a \leq t \leq b$.  To ensure the tangent exists, also suppose that $z'(t)=x'(t)+iy'(t) \neq 0$ for all $a \leq t \leq b$.  We see in Figure \ref{7.1} that the tangent line to the curve at $z_{0}=z(t)$ has the direction of $z'(t)$.  Furthermore, we can determine the direction of $z'(t)$ by $\text{arg}(z'(t))$.

\begin{figure}[H]
\vspace{-.5in}
\centering   
\hspace{.2in}
\begin{minipage}[t]{0.38\textwidth}
\includegraphics[width=\textwidth]{ca6-1-01}\vspace{-.2in}
\caption{The direction of the tangent line at $z(t)$ is given by $\text{arg}(z'(t))$.} \label{7.1}
\end{minipage} 
\hfill
\centering   
\begin{minipage}[t]{0.33\textwidth}
\includegraphics[width=\textwidth]{ca6-1-02}\vspace{-.2in}
\caption{The curves $\gamma_{1}$ and $\gamma_{2}$ intersect at angle $\alpha$.} \label{7.2}
\end{minipage} 
\hspace{.2in}
\vspace{-.2in}
\end{figure}

We let $z_{0}$ be a point in the $z$-plane and consider two smooth paths $\gamma_{1}$ and $\gamma_{2}$ that intersect at $z_{0}$.  Let $L_{1}$ and $L_{2}$ be the tangent lines to $\gamma_{1}$ and $\gamma_{2}$, respectively, at $z_{0}$.  The curves $\gamma_{1}$ and $\gamma_{2}$ intersect at angle $\alpha$ at $z_{0}$ if $L_{1}$ and $L_{2}$ intersect at angle $\alpha$ at $z_{0}$.  This is demonstrated in Figure \ref{7.2}.  

\begin{figure}[H]
\vspace{-1.5in}
\centering   \begin{minipage}[t]{0.74\textwidth}
\includegraphics[width=\textwidth]{ca6-1-03}\vspace{-2.0in}
\caption{A linear mapping $f(z)=az+b$ $(a\neq 0)$ 
rotates by an angle $\arg(a)$, dilates by a factor $|a|$, and translates by $b$.  In particular, it preserves angles between curves.} \label{7.3} 
\end{minipage} 
\vspace{-1.2in}
\end{figure}
  
To begin our discussion on conformal mappings, we will first explore a simple example.  Consider the linear mapping $f(z)=az+b$ where $a,b \in \mathbb{C}$ and $a \neq 0$.  Under the usual interpretation, $f$ takes points in the $z$-plane and maps them to points in $w$-plane.  The map $f$ rotates points in $z$-plane by the fixed angle $\text{arg}(a)$, dilates points by a factor of $|a|$, and translates points by $b$.  If our smooth paths $\gamma_{1}$ and $\gamma_{2}$ intersect at angle $\alpha$ at $z_{0}$ in the $z$-plane, then the images of these curves under $f$ are two paths in the $w$-plane that intersect at $w_{0}=f(z_{0})$.  It is important to note that the angle at which the images of these curves intersect in the $w$-plane is also $\alpha$ because each curve is rotated by the same angle $\text{arg}(a)$ under $f$.  Moreover, $f$ preserves the orientation of $\gamma_{1}$ with respect to $\gamma_{2}$.  
\begin{definition} Let $\Omega$ be a region, an open and connected set.  A mapping $f:\Omega \rightarrow \mathbb{C}$ is said to be conformal at a point $z_{0}$ if it preserves angles and orientation between curves intersecting at $z_{0}$.
\end{definition}

To motivate our first theorem, we make the following observation.  Suppose $f$ is analytic at $z_{0}$ and $f'(z_{0}) \neq 0$.  It follows that $f$ is approximately linear in a neighborhood of $z_{0}$.  That is, 
\begin{equation*}
f(z)=f(z_{0})+f'(z_{0})(z-z_{0})+\varepsilon(z)(z-z_{0}),
\end{equation*}
where $\varepsilon (z) \rightarrow 0$ as $z \rightarrow z_{0}$.  We observe that $f(z)  \approx f'(z_{0})z+b$ where \linebreak $b=-f'(z_{0})z_{0}+f(z_{0})$ is a constant near $z_{0}$.  We see that $f$ is approximately linear, which suggests that $f$ is conformal at $z_{0}$.  The map $f$ rotates by angle $\theta=\text{arg}(f'(z_{0}))$, scales by a factor of $|f'(z_{0})|$, and translates by $b=-f'(z_{0})z_{0}+f(z_{0})$.  

\begin{example} 
We determine the angle of rotation $\alpha=\text{arg}(f'(z))$ and the dilation factor $|f'(z)|$ of the given mappings at the indicated points.
\begin{enumerate}[(a)]
\item $\text{Log}(z)$ at $z=1,i,-i$ 
\item $\dfrac{1+e^{z}}{1-e^{z}}$ at $z=1, i\pi, \ln(3)+2i$
\end{enumerate}
\textbf{Solution.} 
\begin{enumerate}[(a)]
\item Notice that $f(z)=\text{Log}(z)$ is analytic on its domain, which is the complex plane with the principal branch cut.  Note $f'(z)=\frac{1}{z} \neq 0$ for all $z \in \mathbb{C}$.  When $z=1$, $\alpha=\text{arg}(f'(1))=\text{arg}(1)=0+2\pi k=2\pi k$ where $k \in \mathbb{Z}$ and we have a dilation factor of $|f'(1)|=|1|=1$.  When $z=i$, $\alpha=\text{arg}(f'(i))=\text{arg}(f'(-i))=-\frac{\pi}{2}+2\pi k$ for $k \in \mathbb{Z}$ and we have a dilation factor of $|f'(i)|=|-i|=1$.  When $z=-i$, $\alpha=\text{arg}(f'(-i))=\text{arg}(i)=\frac{\pi}{2}+2\pi k$ for $k \in \mathbb{Z}$ and we have a dilation factor of $|f'(-i)|=|i|=1$.
\item Notice that $f(z)=\frac{1+e^{z}}{1-e^{z}}$ is analytic for $\mathbb{C} \backslash \{2k\pi i, k \in \mathbb{Z}\}$ and $f'(z)=\frac{2e^{z}}{(1-e^{z})^{2}}\neq 0$.  When $z=1$, $\alpha=\text{arg}(f'(1))=\text{arg}\left(\frac{2e}{(1-e)^{2}}\right)=0+2\pi k=2\pi k$, where $k \in \mathbb{Z}$, and we have a dilation factor of $|f'(1)|=\left|\frac{2e}{(1-e)^{2}}\right|=\frac{2e}{(1-e)^{2}}$.  \\ When $z=i\pi$, $\alpha=\text{arg}(f'(i\pi))=\text{arg}\left(\frac{2e^{i\pi}}{1-e^{i\pi}}\right)=\text{arg}(-1)=\pi+2\pi k$ where $k \in \mathbb{Z}$.  We have a dilation factor of $|f'(i\pi)|=|-1|=1$.  When $z=\ln(3)+2i$, $\alpha=\text{arg}(f'(\ln(3)+2i))$.  To calculate $f'(\ln(3)+2i)$, 
\begin{align*}
f'(\ln(3)+2i)&=\frac{6e^{2i}}{(1-3e^{2i})^{2}} \\
&=\frac{6e^{2i}}{(1-3e^{2i})^{2}}\frac{(1-3e^{-2i})^{2}}{(1-3e^{-2i})^{2}} \\
&=\frac{6e^{2i}(1-6e^{-2i}+9e^{-4i})}{(10-6\cos(2))^{2}}. 
\end{align*} 
Now, 
\begin{align*}
\text{arg}\left(\frac{6e^{2i}(1-6e^{-2i}+9e^{-4i})}{(10-6\cos(2))^{2}}\right)&=\text{arg}\left(\frac{6}{(10-6\cos(2))^{2}}\right)+ \\
& \hspace{0.5cm} \text{arg}(e^{2i}(1-6e^{-2i}+9e^{-4i})) \\
&=0+2\pi l+\text{arg}(\cos(2)+i\sin(2)-6+ \\
& \hspace{0.5cm} 9\cos(2)-9i\sin(2)) \\
&=\text{arg}(10\cos(2)-6-8i\sin(2))+2\pi l \\
&=\arctan\left(\frac{-4\sin(2)}{5\cos(2)-3}\right)+2\pi k,
\end{align*} 
where $l, k \in \mathbb{Z}$.  
Thus, $\alpha=\arctan\left(\frac{-4\sin(2)}{5\cos(2)-3}\right)+2\pi k$, $k \in \mathbb{Z}$.  We have a dilation factor of 
\begin{align*}
|f'(\ln(3)+2)|&=\left|\frac{6e^{2i}(1-6e^{-2i}+9e^{-4i})}{(10-6\cos(2))^{2}}\right| \\
&=\frac{6\sqrt{100\cos^{2}(2)-120\cos(2)+36+64\sin^{2}(2)}}{(10-6\cos(2))^{2}}.
\end{align*}
\end{enumerate}
\end{example} 

\begin{theorem}[\textit{Conformal Property}] \label{conformal theorem}
Suppose that $f$ is analytic in a neighborhood of a point $z_{0} \in \mathbb{C}$ and suppose that $f'(z_{0}) \neq 0$.  Then $f$ is conformal at $z_{0}$.  
\end{theorem}
\begin{proof}
Let $\gamma$ be any smooth path through $z_{0}$, parametrized by $\gamma(t)=x(t)+iy(t)$ with $\gamma(t_{0})=z_{0}$ and $\gamma'(t_{0}) \neq 0$.  The image of $\gamma$ under $f$ is a path parametrized by $f(\gamma(t))$ that passes through $w_{0}=f(z_{0})$ in the $w$-plane.  Since $\gamma'(t_{0}) \neq 0$, the direction of the tangent line to $\gamma$ at $z_{0}$ is $\text{arg}(\gamma'(t_{0}))$.  By the chain rule,
\begin{equation*}
(f \circ \gamma)'(t_{0})=f'(\gamma(t_{0}))\gamma'(t_{0}) \neq 0
\end{equation*}
because we assumed $f'(z_{0})=f'(\gamma(t_{0})) \neq 0$ and $\gamma'(t_{0}) \neq 0$.  The direction the tangent line to $f(\gamma(t))$ at $w_{0}=f(z_{0})$ is 
\begin{equation*}
\text{arg}((f \circ \gamma)'(t_{0}))=\text{arg}(f'(\gamma(t_{0}))\gamma'(t_{0}))=\text{arg}(f'(\gamma(t_{0})))+\text{arg}(\gamma'(t_{0})).
\end{equation*}
We assumed that $\gamma(t_{0})=z_{0}$, so this equation yields 
\begin{equation*}
\text{arg}((f \circ \gamma)'(t_{0}))=\text{arg}(f'(z_{0}))+\text{arg}(\gamma'(t_{0})).
\end{equation*}
Let $\gamma_{1}$ and $\gamma_{2}$ be two curves with $\gamma_{1}(t_{0})=\gamma_{2}(t_{0})=z_{0}.$  We have the following two equations:
\begin{equation*}
\text{arg}((f \circ \gamma_{1})'(t_{0}))=\text{arg}(f'(z_{0}))+\text{arg}(\gamma_{1}'(t_{0}))
\end{equation*} 
and 
\begin{equation*}
\text{arg}((f \circ \gamma_{2})'(t_{0}))=\text{arg}(f'(z_{0}))+\text{arg}(\gamma_{2}'(t_{0})).
\end{equation*} 
Subtracting these two equations gives us
\begin{equation*}
\text{arg}((f \circ \gamma_{1})'(t_{0}))-\text{arg}(f \circ \gamma_{2})'(t_{0}))=\text{arg}(\gamma_{1}'(t_{0}))-\text{arg}(\gamma_{2}'(t_{0})).
\end{equation*}
We see that the oriented angle of the intersection between any two tangent lines of curves intersecting at $z_{0}$ is the same as that for the images of these curves under the mapping $f$.  Furthermore, $f$ is a conformal mapping by definition.
\hfill\end{proof}
The following examples demonstrate the importance of Theorem \ref{conformal theorem} and its applications.  
\begin{example}
We determine where the given mappings are conformal.
\begin{enumerate}[(a)]
\item $f(z)=\dfrac{\sin(z)}{e^{z}}$
\item $f(z)=\dfrac{e^{z}+1}{e^{z}-1}$ 
\item $f(z)=\dfrac{z+1}{z+i}+\dfrac{2}{z}$
\end{enumerate} 
\textbf{Solution.}
\begin{enumerate}[(a)]
\item Notice for $f(z)=\frac{\sin(z)}{e^{z}}$ that $f'(z)=\frac{\cos(z)-\sin(z)}{e^{z}}$.  The fact that $\sin(z)$ and $e^{z}$ are entire and $e^{z} \neq 0$ implies $f(z)=\frac{\sin(z)}{e^{z}}$ is also entire.  Notice that $f'=0$ when $\cos(z)-\sin(z)=0$.  This occurs when $z=\frac{\pi}{4}+k\pi$ for $k \in \mathbb{Z}$.  Thus, $f$ is conformal for all $z \neq \frac{\pi}{4}+k\pi$ where $k \in \mathbb{Z}$ by Theorem \ref{conformal theorem}. 
\item Notice for $f(z)=\frac{e^{z}+1}{e^{z}-1}$ that $f'(z)=\frac{-2e^{z}}{(e^{z}-1)^{2}}$. The fact that $e^{z}+1$ and $e^{z}-1$ are entire and $e^{z}-1=0$ for $z=2k\pi i$ where $k \in \mathbb{Z}$ implies that $f(z)=\frac{e^{z}+1}{e^{z}-1}$ is analytic for all $z \neq 2k\pi i$ for $k \in \mathbb{Z}$.  Notice that $f' \neq 0$ because $e^{z} \neq 0$.  Thus, $f$ is conformal for all $z \neq 2k\pi i$ where $k \in \mathbb{Z}$ by Theorem \ref{conformal theorem}.
\item  Notice for $f(z)=\frac{z+1}{z+i}+\frac{2}{z}$ that $f'(z)=\frac{i-1}{(z+i)^{2}}-\frac{2}{z^{2}}$.  We know that $\frac{z+1}{z+i}$ is analytic for $\mathbb{C}\backslash\{-i\}$ and $\frac{2}{z}$ is analytic for $\mathbb{C}\backslash\{0\}$ and so $f$ is analytic for $\mathbb{C} \backslash \{-i,0\}$.  To determine when $f'=0$, it remains find $z$ such that $\frac{i-1}{(z+i)^{2}}=\frac{2}{z^{2}}$.  In other words, we will solve $(i-3)z^{2}-4iz+2=0$.  Using the quadratic formula, we have that 
\begin{align*}
z&=\frac{4i \pm \sqrt{(-4i)^{2}-4(i-3)(2)}}{2(i-3)} \\
&=\frac{4i \pm \sqrt{8-8i}}{2i-6}.
\end{align*}
Next, we calculate $\sqrt{8-8i}$.  We know that $r=8\sqrt{2}$ and $\text{Arg}(8-8i)=-\frac{\pi}{4}$. Thus, our two roots are 
\begin{align*}
w_{0}&=2\sqrt{2}\sqrt[4]{2}\left(\cos\left(-\frac{\pi}{8}\right)+i\sin\left(-\frac{\pi}{8}\right)\right) \\
w_{1}&=2\sqrt{2}\sqrt[4]{2}\left(\cos\left(\frac{7\pi}{8}\right)+i\sin\left(\frac{7\pi}{8}\right)\right) \\
&=2\sqrt{2}\sqrt[4]{2}\left(\cos\left(-\frac{\pi}{8}+\pi\right)+i\sin\left(-\frac{\pi}{8}+\pi\right)\right) \\
&=-w_{0}.
\end{align*}
Therefore, $z=\frac{4i+w_{0}}{2i-6}$ or $z=\frac{4i+w_{1}}{2i-6}$ where $w_{0}$ and $w_{1}$ are as above.  We conclude then that $f$ is conformal for all $z\neq -i, 0,\frac{4i+w_{0}}{2i-6},\frac{4i+w_{1}}{2i-6}$ by Theorem \ref{conformal theorem}.
\end{enumerate}
\end{example}

\begin{example}
We consider the orthogonal lines $x=a$ and $y=b$, where $a,b \in \mathbb{R}$.  We determine their images under the given mappings.  We also determine the values of $a$ and $b$ such that the images are orthogonal at the point of intersection $f(a+ib)$ and verify by computing the angle between the image curves at the point $f(a+ib)$.
\begin{enumerate}[(a)]
\item $f(z)=(1+i)z$ 
\item $f(z)=\dfrac{z+1}{z-1}$
\end{enumerate}
\textbf{Solution.}
\begin{enumerate}[(a)]
\item Notice that $f(z)=(1+i)z$ is analytic and $f'(z)=1+i \neq 0$.  This implies that $f$ is conformal for all $\mathbb{C}$.  The image of $x=a$ ($z=a+iy$ for $y \in \mathbb{R}$) is $f(a+iy)=(1+i)(a+iy)=(a-y)+i(a+y)$.  The image of $y=b$ ($z=x+ib$ for $x \in \mathbb{R}$) is $f(x+ib)=(1+i)(x+ib)=(x-b)+i(x+b)$.  When $x=a$, we have $(a-y,a+y)$ by taking the real and imaginary parts of $f(a+iy)$.  Likewise for $y=b$ we have $(x-b,x+b)$.  The tangent vectors of these curves are $(-1,1)$ and $(1,1)$.  Notice that $(-1,1) \cdot (1,1)=0$ and so the images of these two lines are indeed orthogonal. 
\item Notice that $f(z)=\frac{z+1}{z-1}$ is analytic for all $z \in \mathbb{C} \backslash \{1\}$ and $f'(z)=\frac{-2}{(z-1)^{2}}\neq 0$.  This implies that $f$ is conformal for all $z \neq 1$.  Therefore, the images of $x=a$ and $y=b$ should be orthogonal at every value except $x=1$.  Notice that 
\begin{equation*}
f(z)=\frac{z+1}{z-1}=\frac{x^{2}+y^{2}-1}{(x-1)^{2}+y^{2}}+i\left(\frac{-2y}{(x-1)^{2}+y^{2}}\right).
\end{equation*}
When $x=a$, we have the vector $\left(\frac{a^{2}+y^{2}-1}{(a-1)^{2}+y^{2}},\frac{-2y}{(a-1)^{2}+y^{2}}\right)$ and so the tangent vector at the point $(a,b)$ is $\left(\frac{4b-4ab}{((a-1)^{2}+b^{2})^{2}},\frac{-2a^{2}+4a-2+2b^{2}}{((a-1)^{2}+b^{2})^{2}}\right)$.  When $y=b$, we have the vector $\left(\frac{x^{2}+b^{2}-1}{(x-1)^{2}+y^{2}},\frac{-2b}{(x-1)^{2}+b^{2}}\right)$ and so the tangent vector at the point $(a,b)$ is $\left(\frac{-2a^{2}+4a+2b^{2}-2}{((a-1)^{2}+b^{2})^{2}},\frac{4ba-4b}{((a-1)^{2}+b^{2})^{2}}\right)$.  Notice that 
dot product between these two tangent vectors is zero
%\begin{equation*}
%\left(\frac{4b-4ab}{((a-1)^{2}+b^{2})^{2}},\frac{-2a^{2}+4a-2+2b^{2}}{((a-1)^{2}+b^{2})^{2}}\right) \cdot \left(\frac{-2a^{2}+4a+2b^{2}-2}{((a-1)^{2}+b^{2})^{2}},\frac{4ba-4b}{((a-1)^{2}+b^{2})^{2}}\right)=0
%\end{equation*}
so the images of these two curves are indeed orthogonal.
\end{enumerate}
\end{example}
\section*{Boundary Behavior}
When solving boundary value problems, we would like to know if the boundary of a region $\Omega$  is mapped onto the boundary of $f[\Omega]$ under a map $f$.  Conformal mappings are useful for solving boundary value problems on irregular regions by mapping them onto regions for which the particular boundary value problem can be easily solved.  However, under what conditions will the boundary of $\Omega$ be mapped onto the boundary of $f[\Omega]$?  The following example illustrates that the property of being conformal alone is not sufficient.    

\begin{example}\label{6.1Ex1}
Consider $f(z)=z^2$ and $\Omega=\{z=re^{i\theta}: \frac{1}{2}<r<1,-\pi< \theta< \pi\}$.  Here $f$ is analytic and $f'(z)=2z \neq 0$ for all $z$ in $\Omega$.  By Theorem \ref{conformal theorem}, $f$ is conformal for all $z \in \Omega$.  Using DeMoivre's identity, we see that 
\begin{equation*}
f[\Omega]=\{w=re^{i\theta}: \frac{1}{4}<r<1, \,
0\leq \theta\leq 2\pi\}
\end{equation*}

\begin{figure}[H]
\vspace{-2.15in}
  \centering   \begin{minipage}[t]{0.75\textwidth}
    \includegraphics[width=\textwidth]{ca6-1-04}\vspace{-2.0in}
   \caption{The boundary of $\Omega$ is not mapped to boundary of $f[\Omega]$ under $f$.}
   \end{minipage} 
\vspace{-1.75in}
  \end{figure}

Although $f(z)=z^2$ is conformal in $\Omega$, the interval $(-1,-\frac{1}{2})$ on the boundary of $\Omega$ is mapped to the interval $(\frac{1}{4},1)$ in the interior of $f[\Omega]$.  We conclude that the boundary of $\Omega$ is not mapped onto the boundary of $f[\Omega]$ under the map $f$.  
\end{example}
The condition that $f' \neq 0$ for all $z \in \Omega$ ensures that $f$ is one-to-one locally at $z \in \Omega$ by the inverse function theorem \cite{asmar}.  However, the function may fail to be one-to-one on the entire region $\Omega$.  This is exactly what happened in Example \ref{6.1Ex1}, so this leads us to believe that the property of being one-to-one is necessary for a function $f$ to map boundary points to boundary points.  We have the result that if $f$ is analytic and nonconstant on a region $\Omega$, then $f[\Omega]$ is a region \cite{asmar}.  This implies all the points in $\Omega$ are mapped to the open connected set $f[\Omega]$.  It is possible that $f$ is not defined on the boundary of $\Omega$, so we will introduce the following definitions to describe how $f$ maps the boundary of $\Omega$. 
\begin{definition}
Let $f$ be a mapping from a region $\Omega$ to the complex plane. 
\begin{enumerate}[(i)]
\item We say that $f$ {\bf maps the boundary of $\Omega$ to the boundary 
of $f[\Omega]$} if the following condition holds: If $\{z_n\}_{n=1}^\infty$ is any sequence in $\Omega$ converging to $\alpha_0$ on the boundary of $\Omega$ and $\beta$ is any point in $f[\Omega]$, then $\{f(z_n)\}_{n=1}^\infty$ does not converge to $\beta$.  So if $\{f(z_n)\}_{n=1}^\infty$ converges, it must converge to a boundary point of $f[\Omega]$ or to infinity.  If the region $\Omega$ is unbounded, we allow $\alpha_0$ to be $\infty$.
\item We say that $f$ {\bf maps the boundary of $\Omega$ onto the boundary of $f[\Omega]$} if $f$ maps the boundary of $\Omega$ to the boundary of $f[\Omega]$ and, additionally, for every point $w_0$ on the boundary of $f[\Omega]$ there is a sequence $\{z_n\}_{n=1}^{\infty}$ in $\Omega$ with $z_n\rightarrow\alpha_0$, $\alpha_0$ a boundary point of $\Omega$, such that $f(z_n)\rightarrow w_0$.
\end{enumerate} 
\end{definition}

There are cases where $\{f(z_{n})\}_{n=1}^{\infty}$ does not converge, but $\{z_{n}\}_{n=1}^{\infty}$ converges to a point on the boundary of a region.  We will see this in the next example.  
\begin{example}
Let $\Omega$ be the slit plane, $\mathbb{C} \backslash \{z:z \leq 0\}$, and $f(z)=\text{Log}(z)$.  We want to show that $z_{n}=-1+i\frac{(-1)^{n}}{n} (n=1,2,\ldots)$ converges to $-1$ on the boundary of $\Omega$.  To show this, notice that for $n=2k$, then $z_{n}=z_{2k}=-1+\frac{i}{2k}$, which converges to $-1$ as $k \rightarrow \infty$.  Also, notice that for $n=2k+1$, then $z_{n}=z_{2k+1}=-1-\frac{i}{2k+1}$, which converges to $-1$ as $k \rightarrow \infty$.  Thus, we conclude that $z_{n} \rightarrow -1$ as $n \rightarrow \infty$ where $-1$ is on the boundary of $\Omega$.  Now, we will show that $\{f(z_{n})\}_{n=1}^{\infty}$ does not converge.  Notice that 
\begin{align*}
\text{Log}(z_{n})&=\ln\left|-1+i\frac{(-1)^{n}}{n}\right|+i\text{Arg}\left(-1+i\frac{(-1)^{n}}{n}\right) \\ 
&=\ln\left|\sqrt{1+\frac{1}{n^{2}}}\right|+i\text{Arg}\left(-1+i\frac{(-1)^{n}}{n}\right).
\end{align*}
When $n=2k$, $z_{2k} \rightarrow -1$ from the upper-plane implies that $\text{Arg}(z_{2k}) \rightarrow \pi$ as $k \rightarrow \infty$.  Thus, $\text{Log}(z_{2k}) \rightarrow \ln(1)+i(\pi)=i\pi$ as $k \rightarrow \infty$.  When $n=2k+1$, $z_{2k} \rightarrow -1$ from the lower-plane implies that $\text{Arg}(z_{2k+1})\rightarrow -\pi$ as $k \rightarrow \infty$.  Thus, $\text{Log}(z_{2k+1}) \rightarrow \ln(1)+i(-\pi)=-\pi$ as $k \rightarrow \infty$.  Since $\text{Log}(z_{2k}) \rightarrow \pi$ and $\text{Log}(z_{2k+1}) \rightarrow -\pi$ as $k \rightarrow \infty$, we conclude that $\{\text{Log}(z_{n})\}_{n=1}^{\infty}$ does not converge.  
\end{example}
\begin{theorem} \label{boundary}
Suppose that $f$ is analytic and one-to-one on a region $\Omega$.  Then, $f$ maps the boundary of $\Omega$ onto the boundary of $f[\Omega]$.  
\end{theorem}
\begin{proof} Suppose that $f$ is analytic and one-to-one on a region $\Omega$.  First, we will show that $f$ maps the boundary of $\Omega$ to the boundary of $f[\Omega]$.  The fact that $f$ is one-to-one and nonconstant implies that $f[\Omega]$ is a region.  Let $\beta$ be a point in $f[\Omega]$ and suppose, to obtain a contradiction, that $\{z_{n}\}_{n=1}^{\infty}$ is a sequence in $\Omega$ converging to a boundary point $\alpha_{0}$ of $\Omega$ with $f(z_{n}) \rightarrow \beta$.  Notice that $f^{-1}$ exists and is analytic as $f$ is analytic and one-to-one.  Moreover, $f^{-1}$ is continuous and so 
\begin{equation*}
\lim_{n \rightarrow \infty} z_{n}=\lim_{n \rightarrow \infty} f^{-1}(f(z))=f^{-1}(\lim_{n \rightarrow \infty} f(z_{n}))=f^{-1}(\beta).
\end{equation*}
Note that $\beta$ being a point in $f[\Omega]$ implies that $f^{-1}(\beta)$ is a point in $\Omega$.  Therefore, we see that $\{z_{n}\}_{n=1}^{\infty}$ converges to an interior point of $\Omega$, a contradiction.  We conclude that if $\{z_{n}\}_{n=1}^{\infty}$ converges to a point $\alpha_{0}$ on the boundary of $\Omega$, then $\{f(z_{n})\}_{n=1}^{\infty}$ cannot converge to $\beta$.  Second, we will show that $f$ maps the boundary of $\Omega$ onto the boundary of $f[\Omega]$.  To do so, let $w_{0}$ be a point on the boundary of $f[\Omega]$.  There exist points $w_{n}=f(z_{n})$ such that $z_{n}$ is in $\Omega$ and $w_{n} \rightarrow w_{0}$.  Two cases arise.  First, suppose $\{z_{n}\}_{n=1}^{\infty}$ is an unbounded sequence.  There exists a subsequence $\{z_{n_{j}}\}_{j=1}^{\infty}$ that tends to $\infty$, which is a point on the boundary of $\Omega$, if $\Omega$ is an unbounded region.  Moreover, $f(z_{n_{j}}) \rightarrow w_{0}$.  Second, we consider the case in which $\{z_{n}\}_{n=1}^{\infty}$ is a bounded sequence.  Using Bolzano-Weierstrass, we can find a subsequence $\{z_{n_{j}}\}_{j=1}^{\infty}$ that converges to a point $\alpha_{0}$ in $\overline{\Omega}$, the closure of $\Omega$.  If $\alpha_{0}$ were in $\Omega$, then we would have $f(z_{n_{j}}) \rightarrow f(\alpha_{0})$ as $f$ is continuous.  We also know that $f(z_{n_{j}}) \rightarrow w_{0}$, so we would have $f(\alpha_{0})=w_{0}$.  This produces a contradiction as $f(\alpha_{0})$ would lie in $f[\Omega]$ and $w_{0}$ lies on the boundary of $f[\Omega]$.  Thus, $\alpha_{0}$ lies on the boundary of $\Omega$.  
\hfill\end{proof}
\begin{corollary}
Suppose that $f$ is analytic and one-to-one in a region $\Omega$.  Suppose that $f$ extends continuously to $\Omega$.  Then for any point $\alpha$ of the boundary of $\Omega$ the point $f(\alpha)$ lies on the boundary of $f[\Omega]$.  
\end{corollary}
\begin{proof}
Let $\{z_{n}\}_{n=1}^{\infty}$ be a sequence in $\Omega$ with $z_{n} \rightarrow \alpha$ where $\alpha$ is on the boundary of $\Omega$.  By the continuity of $f$, we have that $f(z_{n}) \rightarrow f(\alpha)$.  We know that $f(z_{n})$ does not converge to an interior point of $\Omega$ by Theorem \ref{boundary}, so $f(\alpha)$ must be a boundary point of $f[\Omega]$.  
\hfill\end{proof}
These results about boundary behavior are particularly important for solving Dirichlet problems, but we may also use these results when constructing mappings from regions to regions.  We will now discuss particular types of conformal mappings, linear fractional transformations and Schwarz-Christoffel transformations, in the following two chapters.
\chapter{Linear Fractional Transformations}
We are mostly interested in solving boundary value problems, such as the Dirichlet problem, on the unit disk.  Given a region $\Omega$ that is not the unit disk we will want to find a conformal, one-to-one map that maps $\Omega$ onto the unit disk.  It turns out that the best and most convenient mappings are linear fractional transformations: 
\begin{equation}\label{lft}
\phi(z)=\dfrac{az+b}{cz+d}. \hspace{0.25in} (ad \neq bc)
\end{equation}
The condition $ad \neq bc$ ensures that $\phi'(z)=\dfrac{ad-bc}{(cz+d)^{2}} \neq 0$ for all $z \in \mathbb{C}$, which implies  $\phi$ is not a constant function.  If $c=0$, then $\phi$ is a linear function, which is entire.  If $c \neq 0$, then $\phi$ is analytic for all $z \neq -\frac{d}{c}$ and has a simple pole at $z=-\frac{d}{c}$.

\begin{proposition}
Let $\phi$ be the linear fractional transformation defined in (2.1) where $a,b,c,d \in \mathbb{C}$ with $c \neq 0$ and $ad \neq bc$.  Then, 
\begin{enumerate}[(i)]
\item $\phi$ has a pole at $z=-\frac{d}{c}$ and is $one-to-one$ and conformal on $\mathbb{C} \backslash \{-\frac{d}{c}\}$  and its range is $\mathbb{C} \backslash \{\frac{a}{c}\}$. 
\item The inverse of $\phi$ is the linear fractional transformation from $\mathbb{C} \backslash \{\frac{a}{c}\}$ onto $\mathbb{C} \backslash \{-\frac{d}{c}\}$ given by 
\begin{equation}
\psi(w)=\dfrac{dw-b}{-cw+a}.
\end{equation}
\end{enumerate}
\end{proposition} \label{inverse prop}
\begin{proof}
(i) Let $\phi(z)=\frac{az+b}{cz+d}$ with $a,b,c,d \in \mathbb{C}$, $c \neq 0$, and $ad \neq bc$.  Notice that $\phi$ has a simple pole at $z=-\frac{d}{c}$.  Furthermore, $\phi'(z)=\frac{ad-bc}{(cz+d)^{2}} \neq 0$ for all $z \in \mathbb{C} \backslash \{-\frac{d}{c}\}$.  By Theorem \ref{conformal theorem}, $\phi$ is conformal at $z \neq -\frac{d}{c}$.  To show that $\phi$ is one-to-one, let $z_{1}, z_{2} \in \mathbb{C} \backslash \{-\frac{d}{c}\}$ and suppose $\phi(z_{1})=\phi(z_{2})$.  Using the fact that $\phi(z_{1})=\phi(z_{2})$ and multiplying by $(cz_{1}+d)(cz_{2}+d)$, we obtain 
\begin{equation*}
(ad-bc)(z_{1}-z_{2})=0.
\end{equation*}
We assumed that $ad-bc \neq 0$, so it must be that $z_{1}-z_{2}=0$.  Hence, $z_{1}=z_{2}$ and so $\phi$ is one-to-one.  
To determine the range of $\phi$, let $z^{*} \in \mathbb{C}$.  We want to find a $z \in \mathbb{C}$ such that $\phi(z)=z^{*}$.  Here $z=\frac{-b+z^{*}d}{a-z^{*}c}$ which is not defined for $z^{*}=\frac{a}{c}$.  Furthermore, the range of $\phi$ is $\mathbb{C} \backslash \{\frac{a}{c}\}$.  (ii) Notice that $\phi:\mathbb{C} \backslash \{-\frac{d}{c}\} \rightarrow \mathbb{C} \backslash \{\frac{a}{c}\}$ is bijective and so $\phi$ has an inverse.  Namely, it is defined by 
\begin{equation*}
\psi(w)=\frac{dw-b}{a-wc}.
\end{equation*}
Therefore, we conclude that $(i)$ and $(ii)$ hold. 
\hfill\end{proof}
We now observe that every such linear fractional transformation with $c \neq 0$ is the composition of a linear mapping $w_{1}=cz+d$, $w_{2}=\frac{1}{w_{1}}$, and a linear mapping $w=\frac{a}{c}+(b-\frac{ad}{c})w_{2}$.  Linear mappings map lines to lines and circles to circles.  This is quickly verfied by the the fact that a linear mapping $f(z)=az+b$ rotates by an angle $\text{arg}(a)$, dilates by a factor of $|a|$, and translates by $b$ units.  On the other hand, the inversion map $f(z)=\frac{1}{z}$ maps lines and circles to either lines or circles.  This is more difficult to see, but we will prove the following results about the inversion map.  
\begin{lemma} \label{linescirclesform}
Any circle of the form $(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$ for $r>0$ and any line in the plane can be written in the form 
\begin{equation*}
A(x^{2}+y^{2})+Bx+Cy+D=0,
\end{equation*}
where $B^{2}+C^{2}-4AD>0$.  
\end{lemma}
\begin{proof}
If we a have a circle of the form $(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}$ for $r>0$, we can obtain $(x^{2}+y^{2})-2x_{0}x-2y_{0}y+({x_{0}}^{2}+{y_{0}}^{2}-r^{2})=0$ by expanding the initial terms and grouping the appropriate terms.  We see that $A=1$, $B=-2x_{0}$, $C=-2y_{0}$, and $D={x_{0}}^{2}+{y_{0}}^{2}-r^{2}$.  We know that a line is of the form $y=mx+b$, which implies $y-mx-b=0$.  We see that $A=0$, $B=-m$, $C=1$, and $D=-b$.  
\hfill\end{proof}
\begin{lemma}\label{formsforlinescircles}
Any set of points satisfying $A(x^{2}+y^{2})+Bx+Cy+D=0$ with $B^{2}+C^{2}-4AD>0$ is either a circle or a line depending on whether $A=0$ or $A \neq 0$.  Futhermore, such circles and lines pass through the origin if and only if $D=0$.  
\end{lemma}
\begin{proof}
If $A=0$, then we have $Bx+Cy+D=0$ where $B^{2}+C^{2}>0$.  The inequality $B^{2}+C^{2}>0$ ensure that both $B$ and $C$ cannot be zero at the same time.  Notice that $Bx+Cy+D=0$ is a line.  When $A \neq 0$, then we have $A(x^{2}+y^{2})+Bx+Cy+D=0$.  Completing the square yields $(x+\frac{B}{2A})^{2}+(y+\frac{C}{2A})^{2}=\frac{B^{2}+C^{2}-4AD}{4A^{2}}$.  The fact that $B^{2}+C^{2}-4AD>0$ implies we have a circle with center $(-\frac{B}{2A},-\frac{C}{2A})$ and radius $r=\frac{\sqrt{B^{2}+C^{2}-4AD}}{2A}$.  A line is of the form $Bx+Cy+D=0$ where $B^{2}+C^{2}>0$.  If $D=0$, then we simply have $Bx+Cy=0$, which is a line passing through the origin.  If the line passes through the origin (i.e. $(x,y)=(0,0)$), we have $D=0$.  A circle is of the form $(x+\frac{B}{2A})^{2}+(y+\frac{C}{2A})^{2}=\frac{B^{2}+C^{2}-4AD}{4A^{2}}$.  If $D=0$, this reduces to a circle of the form $(x+\frac{B}{2A})^{2}+(y+\frac{C}{2A})^{2}=\frac{B^{2}+C^{2}}{4A^{2}}$.  Note that the distance between the center $(-\frac{B}{2A},-\frac{C}{2A})$ and the origin is $\frac{\sqrt{B^{2}+C^{2}}}{2A}$, which is precisely the radius of this circle.  Therefore, the circle must pass through the origin.  On the other hand, if $x=y=0$, we have that $\frac{B^{2}+C^{2}}{4A^{2}}=\frac{B^{2}+C^{2}-4AD}{4A^{2}}$.  This implies that $-4AD=0$ and so $D=0$ as $A \neq 0$. 
\hfill\end{proof}
\begin{proposition}
Let $S$ be a set of points $(x,y)$ satisfying $A(x^{2}+y^{2})+Bx+Cy+D=0$ where $B^{2}+C^{2}-4AD>0$.  Then, under the map $f(z)=\frac{1}{z}$, points $(u,v)$ in $f[S]$ satisfy an equation of the same form.  Morever, lines and circles are mapped to lines and circles, with the exception of the origin.  
\end{proposition}
\begin{proof}
Assume $S$ is a set of points $(x,y)$ satisfying $A(x^{2}+y^{2})+Bx+Cy+D=0$ where $B^{2}+C^{2}-4AD>0$.  By Lemma \ref{linescirclesform}, $S$ is either a line or circle.  For any point $z \in S$, we have 
\begin{equation*}
\frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+ y^2}.
\end{equation*}
We conclude that 
\begin{equation*}
f[S]=\left\{u+iv: u=\frac{x}{x^2+ y^2},\ v=\frac{-y }{x^2+ y^2} 
\right\}
\end{equation*}
with $x$ and $y$ satisfying $A(x^2+y^2)+Bx+Cy+D=0$ with  $B^2+C^2-4AD>0$.
We want to prove that $u$ and $v$ satisfy a similar identity with possibly different values of the coefficients.
Notice that 
\begin{align*}
E(u^2+v^2)+Fu+Gv+H&=E \frac{x^2 +y^2}{\big(x^2+ y^2\big)^2}+\frac{Fx}{x^2+ y^2}-\frac{Gy}{x^2+ y^2}+H\\
&=\frac{E}{x^2+ y^2}+\frac{Fx }{x^2+ y^2}-\frac{Gy }{x^2+ y^2}+\frac{H(x^2+y^2)}{x^2+ y^2} \\
&=\frac{1}{x^2+y^2}\big(H(x^2+y^2)+Fx-Gy+E\big).
\end{align*}
If we choose $H=A$, $F=B$, $G=-C$, and $E=D$, then, for all points in $S\backslash\{(0,0)\}$, the last
expression is equal to 0 by assumption.  We conclude that $(u,v)$ satisfy 
\begin{equation*}
D(u^2+v^2)+Bu-Cv+A=0
\end{equation*}
with $B^2+C^2-4AD>0$.  By Lemma \ref{formsforlinescircles}, we conclude that $f[S]$ is either a line or circle.
\hfill\end{proof}
Because linear fractional transformations are a composition of two linear mappings and the inversion map, linear fractional transformations will map lines to either lines or circles and circles to either lines or circles.  We illustrate this in the next example.
\begin{example}
Let $L_{1}$ denote the line through $z_{1}$ and $z_{2}$ and $L_{2}$ the line through $z_{2}$ and $z_{3}$.  We describe the images by $\phi$ of $L_{1}$ and $L_{2}$.
\begin{enumerate}
\item[(a)] $\phi(z)=i\dfrac{1-z}{1+z}, z_{1}=1, z_{2}=0, z_{3}=i$
\item[(b)] $\phi(z)=\dfrac{1+i-2z}{i-iz}, z_{1}=i, z_{2}=1, z_{3}=-i$
\end{enumerate}
\end{example}
\noindent \textbf{Solution.} 
\begin{enumerate}
\item[(a)] Let $L_{1}$ be the line through $z_{1}=1$ and $z_{2}=0$ and $L_{2}$ be the line through $z_{2}=0$ and $z_{3}=i$.  Notice that the $L_{1}$ is perpendicular to $L_{2}$, so we would expect $\phi[L_{1}]$ to be perpendicular to $\phi[L_{2}]$.  First, we consider $\phi[L_{1}]$.  Notice that the point $z=-1$ is on $L_{1}$ and $\phi(-1)=\infty$.  Thus, we know that $\phi[L_{1}]$ is a line.  Since $\phi(1)=0$ and $\phi(0)=i$, we conclude that $\phi[L_{1}]$ is the imaginary axis in the $w$-plane.  Since $\phi[L_{1}]$ is the imaginary axis and $\phi[L_{2}]$ must go through $w_{2}=\phi(0)=i$ and $w_{3}=\phi(i)=1$, we would expect that $\phi[L_{2}]$ is a circle.  In addition to the points $w_{2}$ and $w_{3}$, we have that $\phi(-i)=-1$.  This tells us that $\phi[L_{2}]$ is a circle through the points $w_{2}=i$, $w_{3}=1$, and $w_{4}=-1$.
\item[(b)] Let $L_{1}$ denote the line through the points $z_{1}=i$ and $z_{2}=1$ and $L_{2}$ denote the line through the points $z_{2}=1$ and $z_{3}=-i$.  Notice that $\phi[L_{1}]$ will go through $w_{1}=-i$ and $w_{2}=\infty$ and so we conclude that $\phi[L_{1}]$ is a line through the point $w_{1}=-i$.  To determine the line, we use the point $z=\frac{1+i}{2}$ as a second point.  Notice that $w_{4}=\phi\left(\frac{1+i}{2}\right)=0$.  Therefore, we conclude that $\phi[L_{1}]$ is the imaginary axis in the $w$-plane.  Next, $\phi[L_{2}]$ will go through $w_{2}=\infty$ and $w_{3}=1-2i$, which tells us that $\phi[L_{2}]$ is a line.  To determine the line, we use $z=\frac{1-i}{2}$ as a second point.  Notice that $w_{5}=\phi\left(\frac{1-i}{2}\right)=2-2i$.  Thus, $\phi[L_{2}]$ is a line going through the points $w_{3}=1-2i$ and $w_{5}=2-2i$.   
\end{enumerate} 
We will now address two common mappings that are used to transfer between the unit disk and the upper half-plane.
\begin{example} \label{basic map}
\begin{enumerate}[(a)]
\item We show that the linear fractional transformation 
\begin{equation*}
\phi(z)=i\frac{1-z}{1+z}
\end{equation*}
maps the unit disk onto the upper half-plane.
\item We show that the linear fractional transformation 
\begin{equation*}
\psi(z)=\frac{i-z}{i+z}
\end{equation*}
maps the upper half-plane onto the unit disk. 
\end{enumerate}
\end{example}
\noindent \textbf{Solution.} 
\begin{enumerate}[(a)]
\item The image of the unit circle $C_{1}(0)$ is either a line or a circle.  Computing the images of three points on $C_{1}(0)$ enables us to determine whether or not $\phi[C_{1}(0)]$ is a line or circle in the $w$-plane.  Notice that $\phi(1)=0$, $\phi(i)=1$, and $\phi(-i)=-1$.  These values lie on the $u$-axis of the $w$-plane, which suggests that $\phi[C_{1}(0)]$ is the $u$-axis.  The fact that $\phi$ is one-to-one implies that the boundary of the unit disk is mapped onto the boundary of its image under $\phi$, the $u$-axis, by Theorem \ref{boundary}.  This implies that the image of the unit disk is either the upper half-plane or the lower half-plane.  Observe that $\phi(0)=i$, so we conclude that the unit disk is mapped to the upper half-plane.  Note that $\phi$ has a singularity at $z=-1$, which enables the bounded, unit disk to be mapped onto the unbounded, upper half-plane.  For the point $z=-1$ on $C_{1}(0)$, we see that $\lim_{z \rightarrow -1} \phi(z)=\infty$.  The image of $C_{1}(0)$ is unbounded and thus has to be a line as a circle is bounded.  

\begin{figure}[H]
\vspace{-1.7in}
\centering   \begin{minipage}[t]{0.8\textwidth}
\includegraphics[width=\textwidth]{ca6-2-02}\vspace{-2.0in}
\caption{$\phi(z)=i\frac{1-z}{1+z}$ maps the unit disk to the upper half-plane.}
\end{minipage} 
\vspace{-1.8in}
\end{figure}
  
\item One can see that $\psi$ is the inverse of $\phi$, so $\psi$ will take the upper half-plane to the unit circle.  Alternatively, we can proceed as in part (a).  Note that $\psi(0)=1$, $\psi(1)=i$, and $\psi(-1)=-i$.  Observe the images of three collinear points are not collinear.  Morever, these points all lie on the unit circle, so the image of the $x$-axis under $\psi$ is $C_{1}(0)$.  Since $\psi(i)=0$, we conclude that $\psi$ maps the upper half-plane onto the unit disk.  
\end{enumerate}

We saw in this example that $\lim_{z \rightarrow -1} \phi(z)=\infty$ and $\lim_{z \rightarrow \infty} \psi(z)=-1$.  We will generally say $\phi(z_{0})=\infty$ to mean the limit at the point $z_{0}$ is infinite.  Similarily, if $\lim_{z \rightarrow \infty} \frac{az+b}{cz+d}=\frac{a}{c}$, we will write $\phi(\infty)=\frac{a}{c}$.   
\begin{proposition}[Composition of Mappings]
The composition of any two linear fractional transformations is another linear fractional transformation.  
\end{proposition}
\begin{proof}
Let $\phi_{1}(z)=\dfrac{a_{1}z+b_{1}}{c_{1}z+d_{1}}$ and $\phi_{2}(z)=\dfrac{a_{2}z+b_{2}}{c_{2}z+d_{2}}$.  It follows that
\begin{equation*}
\phi(z)=(\phi_{2} \circ \phi_{1})(z)=\frac{(a_{1}a_{2}+b_{2}c_{1})z+(a_{2}b_{1}+b_{2}d_{1})}{(a_{1}c_{2}+c_{1}d_{2})z+(b_{1}c_{2}+d_{1}d_{2})}.
\end{equation*}
If we let $a=a_{1}a_{2}+b_{2}c_{1}$, $b=a_{2}b_{1}+b_{2}d_{1}$, $c=a_{1}c_{2}+c_{1}d_{2}$, and $d=b_{1}c_{2}+d_{1}d_{2}$, we see that $ad-bc=(a_{1}d_{1}-b_{1}c_{1})(a_{2}d_{2}-b_{2}c_{2}) \neq 0$ as $a_{1}d_{1}-b_{1}c_{1} \neq 0$ and $a_{2}d_{2}-b_{2}c_{2} \neq 0$.  
Thus, $\phi$ is a linear fractional transformation with one pole.  This pole may or may not occur at the same location as the poles of either $\phi_{1}$ or $\phi_{2}$.
\hfill\end{proof}
\begin{example}
We find a linear fractional transformation that maps the disk $B_{2}(-1)$, a disk with radius $2$ and center $-1$, onto the right half-plane. 
\end{example}
\noindent \textbf{Solution.}
We construct a linear fractional transformation $f$ that maps the disk $B_{2}(-1)$ onto the right half-plane through a composition of linear fractional transformations.  Note that that $f_{1}: B_{2}(-1) \rightarrow B_{1}(0)$ defined by $f_{1}(z)=\frac{1}{2}(z+1)$ is a linear mapping.  This map shifts the disk $B_{2}(-1)$ right one unit and shrinks the disk by a factor of two.  By Example \ref{basic map}, $f_{2}(w_{1})=i\frac{1-w_{1}}{1+w_{1}}$ maps the unit circle onto the upper half-plane.  Lastly, the map $f_{3}(w_{2})=-iw_{2}$ maps the upper half-plane onto the right half-plane by rotating the upper half-plane $90^{\circ}$ clockwise.  Note that each of these maps is a linear fractional transformation and so the composition $f(z)=f_{3}(f_{2}(f_{1}(z)))=\frac{1-z}{3+z}$ is also a linear fractional transformation.  Moreover, $f$ maps the disk $B_{2}(-1)$ onto the right half-plane.  This composition is demonstrated in Figure \ref{easyexample}.  

\begin{figure}[H]
\vspace{-3in}
  \centering   \begin{minipage}[t]{0.99\textwidth}
    \includegraphics[width=\textwidth]{ca6-2-03}\vspace{-3in}
   \caption{To map a disk to the half-plane,
it is always advantageous to map the given disk to the unit disk and then use the transformation $\phi$ in Example \ref{basic map}.\label{easyexample}}
   \end{minipage} 
\vspace{-2.6in}
  \end{figure}


Before we explore more examples of similar compositions, we introduce propositions that are useful for explicitly finding linear fractional transformations given information on three distinct points and their images.  

\begin{lemma}[Fixed points] \label{fixedpoints}
A linear fractional transformation $\phi$ can have at most two fixed points in the complex plane, unless $\phi(z)=z$, in which case all points are fixed points.  
\end{lemma}
\begin{proof}
Let $\phi$ be a linear fractional transformation.  We consider all solutions to the equation $z=\dfrac{az+b}{cz+d}$.  Two cases arise.  First, let's assume $c=0$.  Then, we will solve $z=\dfrac{az+b}{d}$.  Here $z=\dfrac{b}{d-a}$.  Next, let's assume $c \neq 0$.  Then, $z=\dfrac{az+b}{cz+d}$ implies that $z(cz+d)=az+b$.  This further implies that $cz^{2}+(d-a)z-b=0$, which has at most two solutions.  In either case, we see that $z$ has at most two solutions.  Therefore, there are at most two fixed points for $\phi$ unless $\phi$ is the identity map, in which case all points are fixed points. 
\hfill\end{proof}
\begin{lemma}[Uniqueness of a linear fractional transformation] \label{uniqueness} 
Let $z_{1}$, $z_{2}$, and $z_{3}$ be three distinct points, and let $w_{1}$, $w_{2}$, and $w_{3}$ be three distinct points (we allow $\infty$).  If there is a linear fractional transformation mapping $z_{j}$ to $w_{j}$ for $j=1,2,3$, then it is unique.  
\end{lemma}
\begin{proof}
Let $f$ and $g$ be two linear fractional transformations mapping $z_{j}$ to $w_{j}$ for $j=1,2,3$.  Notice that $g^{-1}$ exists and is also a linear fractional transformation mapping $w_{j}$ to $z_{j}$ for $j=1,2,3$ by Proposition 2.1.  Notice then that the composition $g^{-1} \circ f$ is also a linear fractional transformation by Proposition 2.3.  Then, we have 
\begin{equation*}
g^{-1}(f(z_{j}))=g^{-1}(w_{j})=z_{j}.
\end{equation*}
Since this is true for all $j=1,2,3$, then we conclude that $g^{-1} \circ f$ has three fixed points.  By Lemma \ref{fixedpoints}, we conclude that $g^{-1}(f(z))=z$ and so $f=g$. 
\hfill\end{proof}
\begin{proposition} \label{functionmaker}
There is a unique linear fractional transformation $w$ that maps three distinct points $z_{1}, z_{2},$ and $z_{3}$ onto three distinct points $w_{1}, w_{2}$, and $w_{3}$.  The mapping $w$ is given implicitly (as a function of $z$) by 
\begin{equation}
\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}.
\end{equation}
\end{proposition}
\begin{proof}
We can see that $w$ given above is indeed a linear fractional transformation if we solve for $w$.  To do so, let $\alpha=\frac{z_{2}-z_{3}}{z_{2}-z_{1}}$ and $\beta=\frac{w_{2}-w_{3}}{w_{2}-w_{1}}$.  Note $\alpha, \beta \neq 0$ as $z_{j}$ and $w_{j}$ are unique for $j=1,2,3$.  Solving for $w$, we obtain
\begin{equation*}
w=\frac{(\alpha w_{3}-\beta w_{1})z+(\beta w_{1}z_{3}-\alpha z_{1}w_{3})}{(\alpha-\beta)z+(\beta z_{3}-\alpha z_{1})}.
\end{equation*}
Let $a=\alpha w_{3}-\beta w_{1}$, $b=\beta w_{1}z_{3}-\alpha z_{1}w_{3}$, $c=\alpha-\beta$, and $d=\beta z_{3}-\alpha z_{1}$.  We have that $a,b,c,d \in \mathbb{C}$ and in fact 
\begin{align*}
ad-bc=\alpha \beta (w_{1}-w_{3})(z_{1}-z_{3}) \neq 0
\end{align*}
by the uniqueness of the points.  
Using our explicit formula for $w$, we see $w(z_{1})=w_{1}$, $w(z_{2})=w_{2}$, and $w(z_{3})=w_{3}$.  Uniqueness follows by Lemma \ref{uniqueness}.  
\hfill\end{proof}
\begin{proposition}\label{somethingmapstoinfinity}
Let $z_{1}, z_{2},$ and $z_{3}$ be three distinct points.  There is a unique linear fractional transformation $w$ that maps $z_{1}$ and $z_{2}$ onto two distinct points $w_{1}$ and $w_{2}$, and maps $z_{3}$ to $\infty$.  The mapping $w$ is given implicitly (as a function of $z$) by 
\begin{equation*}
\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w_{2}-w_{1}}.
\end{equation*}
\end{proposition}
\begin{proof}
We begin by solving 
\begin{equation*}
\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w_{2}-w_{1}}
\end{equation*}
for $w$.  It follows that 
\begin{align*}
w&=\frac{(z-z_{1})(z_{2}-z_{3})(w_{2}-w_{1})}{(z-z_{3})(z_{2}-z_{1})}+w_{1} \\
&=\frac{z[(z_{2}-z_{3})(w_{2}-w_{1})+w_{1}(z_{2}-z_{1})]-z_{1}(z_{2}-z_{3})(w_{2}-w_{1})-w_{1}z_{3}(z_{2}-z_{1})}{z(z_{2}-z_{1})-z_{3}(z_{2}-z_{1})}.
\end{align*}
Notice that $w$ is in the form $w=\dfrac{az+b}{cz+d}$.  Here we have 
\begin{align*}
a&=(z_{2}-z_{3})(w_{2}-w_{1})+w_{1}(z_{2}-z_{1}) \\
b&=-z_{1}(z_{2}-z_{3})(w_{2}-w_{1})-w_{1}z_{3}(z_{2}-z_{1}) \\
c&=z_{2}-z_{1} \\
d&=-z_{3}(z_{2}-z_{1}).
\end{align*}
Notice that $a,b,c,d \in \mathbb{C}$, so we just need to verify that $ad-bc \neq 0$. 
\begin{equation*}
ad-bc=(z_{1}-z_{3})(z_{2}-z_{1})(z_{2}-z_{3})(w_{2}-w_{1}) \neq 0
\end{equation*}
by the uniqueness of the points. 
Next, we verify that $w$ maps $z_{1}$ and $z_{2}$ onto two distinct points $w_{1}$ and $w_{2}$, and maps $z_{3}$ to $\infty$.  When $z=z_{1}$ and $w=w_{1}$, the equation $\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w_{2}-w_{1}}$ results in $0=0$.  When $z=z_{2}$ and $w=w_{2}$, the equation $\frac{z-z_{1}}{z-z_{3}}\frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w_{2}-w_{1}}$ results in $1=1$.  The fact that 
\begin{equation*}
\lim_{z \rightarrow z_{3}} \frac{z-z_{1}}{z-z_{3}} \frac{z_{2}-z_{3}}{z_{2}-z_{1}}=\infty \, \,  \text{and} \, \, \lim_{w \rightarrow \infty} \frac{w-w_{1}}{w_{2}-w_{1}}=\infty
\end{equation*}
tells us that $z_{3}$ maps to $\infty$.  Uniqueness follows by Lemma \ref{uniqueness}.  
\hfill\end{proof}

\begin{proposition}\label{infinitymapsto}
Let $z_{1}$ and $z_{2}$ be two distinct points, and $w_{1}, w_{2},$ and $w_{3}$ be three distinct points.  There is a unique linear fractional transformation $w$ that maps $z_{1}$ to $w_{1}$, $z_{2}$ to $w_{2}$, and $\infty$ to $w_{3}$.  The mapping $w$ is implicitly given (as a function of $z$) by 
\begin{equation*}
\frac{z-z_{1}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}.
\end{equation*}
\end{proposition}
\begin{proof}
We begin by solving 
\begin{equation*}
\frac{z-z_{1}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}
\end{equation*}
for $w$.  Solving for $w$ yields the following equation
\begin{equation*}
w=\frac{-w_{3}z(w_{2}-w_{1})+w_{3}z_{1}(w_{2}-w_{1})+w_{1}(z_{2}-z_{1})(w_{2}-w_{3})}{-z(w_{2}-w_{1})+z_{1}(w_{2}-w_{1})+(z_{2}-z_{1})(w_{2}-w_{3})}.
\end{equation*}
Notice that $w$ is in the form $w=\frac{az+b}{cz+d}$.  Here we have 
\begin{align*}
a&=-w_{3}(w_{2}-w_{1}) \\
b&=w_{3}z_{1}(w_{2}-w_{1})+w_{1}(z_{2}-z_{1})(w_{2}-w_{3}) \\
c&=-(w_{2}-w_{1}) \\
d&=z_{1}(w_{2}-w_{1})+(z_{2}-z_{1})(w_{2}-w_{3}).
\end{align*}
Notice $a,b,c,d \in \mathbb{C}$, so we just need to verify that $ad-bc \neq 0$.  Observe that
\begin{equation*}
ad-bc=(w_{3}-w_{1})(w_{2}-w_{1})(w_{2}-w_{3})(z_{2}-z_{1}) \neq 0
\end{equation*}
by the uniqueness of the points.  Next, we will verify that $w$ maps $z_{1}$ to $w_{1}$, $z_{2}$ to $w_{2}$, and $\infty$ to $w_{3}$.  When $z=z_{1}$ and $w=w_{1}$, then the equation $\frac{z-z_{1}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}$ yields $0=0$.  When $z=z_{2}$ and $w=w_{2}$, then the equation $\frac{z-z_{1}}{z_{2}-z_{1}}=\frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}$ yields $1=1$.  The fact that 
\begin{equation*}
\lim_{z \rightarrow \infty} \frac{z-z_{1}}{z_{2}-z_{1}}=\infty \, \, \text{and} \, \, \lim_{w \rightarrow w_{3}} \frac{w-w_{1}}{w-w_{3}}\frac{w_{2}-w_{3}}{w_{2}-w_{1}}=\infty
\end{equation*}
tells us that $\infty$ maps to $w_{3}$.  Uniqueness follows by Lemma \ref{uniqueness}.  
\hfill\end{proof}
We will now consider applications of these results in the following examples.
\begin{example} \label{lft1}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{1fig}  and derive the given formula for the final composite mapping.  
\begin{figure}[H] 
\vspace{-3in}
\centering   \begin{minipage}[t]{0.99\textwidth}
\includegraphics[width=\textwidth]{ca6-2-13L}\vspace{-3.20in}\caption{Sequence of mappings for Example \ref{lft1}.} \label{1fig} 
\end{minipage} 
\vspace{-3.20in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} 
First, we want to shift our graph down by $1+i$ units so that the circle is centered at the origin.  This mapping is given by $f_{1}(z)=z-1-i$.  Next, we use the inversion map $f_{2}(w_{1})=\frac{1}{w_{1}}$ to map to the interior of the unit circle.  Lastly, $f_{3}(w_{2})=i\frac{1-w_{2}}{1+w_{2}}$ by Example \ref{basic map}.  Furthermore, the composition of these maps is given by
\begin{equation*}
g(z)=f_{3}(f_{2}(f_{1}(z)))=i\frac{z-2-i}{z-i}.
\end{equation*}

\begin{example} \label{lft2}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{2fig} and derive the given formula for the final composite mapping.  
\begin{figure}[H] 
\vspace{-3in}
\centering   \begin{minipage}[t]{0.99\textwidth}
\includegraphics[width=\textwidth]{ca6-2-14}\vspace{-3.2in}\caption{Sequence of mappings for Example \ref{lft2}.} \label{2fig}
\end{minipage} 
\vspace{-3.10in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} 
To construct $f_{1}$, we use that fact that $-2 \mapsto 0$, $0 \mapsto 1$, and $2 \mapsto \infty$.  Using Proposition \ref{functionmaker}, we get that $f_{1}(z)=\frac{-z-2}{z-2}$.  Next, we have that $f_{2}(w_{1})={w_{1}}^{6}$ maps the region in the second picture to the upper half-plane by DeMoivre's identity.  Lastly, $f_{3}(w_{2})=\frac{i-w_{2}}{i+w_{2}}$ by Example \ref{basic map}.  Furthermore, the composition of these maps is given by
\begin{equation*}
g(z)=f_{3}(f_{2}(f_{1}(z)))=\frac{i(z-2)^{6}-(z+2)^{6}}{i(z-2)^{6}+(z+2)^{6}}.
\end{equation*}


\begin{example}\label{lft3}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{u} and derive the given formula for the final composite mapping.  
\begin{figure}[H] 
\vspace{-3.0in}
\centering   \begin{minipage}[t]{0.99\textwidth}
\includegraphics[width=\textwidth]{ca6-2-18}\vspace{-3.15in}\caption{Sequence of mappings for Example \ref{lft3}.} \label{u}
\end{minipage} 
\vspace{-3.00in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} To construct $f_{1}$, notice that $2i \mapsto -\frac{i}{2}$, $-2 \mapsto -\frac{1}{2}$, and $0 \mapsto \infty$.  Using Proposition \ref{functionmaker}, we get that $f_{1}(z)=\frac{1}{z}$.  Next, we have $f_{2}(w_{1})=w_{1}+\frac{1+i}{2}$ as a shift of $\frac{1+i}{2}$ was preformed to transition between the second and third regions.  Lastly, $f_{3}(w_{2})=(-iw_{2})^{\frac{2}{3}}$ by DeMoivre's identity.  Furthermore, the composition of these maps is given by 
\begin{equation*}
g(z)=f_{3}(f_{2}(f_{1}(z)))=\left(\frac{-i}{z}+\frac{1-i}{2}\right)^{\frac{2}{3}}.  
\end{equation*}
\begin{example}\label{lft4}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{4} and derive the given formula for the final composite mapping.  
\begin{figure}[H]
\vspace{-2.5in}
\centering   \begin{minipage}[t]{0.9\textwidth}
\includegraphics[width=\textwidth]{ca6-2-19}\vspace{-2.65in}\caption{Sequence of mappings for Example \ref{lft4}.} \label{4}
\end{minipage} 
\vspace{-2.8in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} To construct $f_{1}$, notice that $-2 \mapsto \frac{3i}{2}, -1 \mapsto i,$ and $1 \mapsto 0$.  Using Proposition \ref{functionmaker}, we obtain $f_{1}(z)=i\frac{1-z}{2}$.  To construct $f_{2}$, notice that $i \mapsto 0$, $0 \mapsto 1$, and $\infty \mapsto -1$.  Using Proposition \ref{infinitymapsto}, we obtain $f_{2}(w_{1})=-\frac{w_{1}-i}{w_{1}+i}$.  Thus, the composition of these functions is given by $g(z)=f_{2}(f_{1}(z))=\frac{z+1}{-z+3}$.  

\begin{example} \label{lft5}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{5fig} and derive the given formula for the final composite mapping.  
\begin{figure}[H] 
\vspace{-2.5in}
\centering   \begin{minipage}[t]{0.99\textwidth}
\includegraphics[width=\textwidth ]{ca6-2-20}\vspace{-3.0in}\caption{Sequence of mappings for Example \ref{lft5}.} \label{5fig}
\end{minipage} 
\vspace{-3.0in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} First, $f_{1}(z)=\frac{1}{4}z^{2}$ maps the quarter circle with radius two to the upper half-circle with radius one as a result of DeMoivre's identity and a shrink.  To construct $f_{2}$, notice that $1 \mapsto 0$, $0 \mapsto i$, and $-1 \mapsto \infty$.  By Proposition \ref{somethingmapstoinfinity} , we have have $f_{2}(w_{1})=-i\frac{w_{1}-1}{w_{1}+1}$.  The composition of these maps is given by
\begin{equation*}
g(z)=f_{2}(f_{1}(z))=i\left(\frac{-z^{2}+4}{z^{2}+4}\right).
\end{equation*}

\begin{example} \label{lft6}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{6fig} and derive the given formula for the final composite mapping.  
\begin{figure}[H]
\vspace{-2.7in}
\centering   \begin{minipage}[t]{0.95\textwidth}
\includegraphics[width=\textwidth]{ca6-2-23}\vspace{-2.8in}\caption{Sequence of mappings for Example \ref{lft6}.} \label{6fig}
\end{minipage} 
\vspace{-2.8in}
\end{figure}
\end{example}
\noindent \textbf{Solution.} We have $f_{1}(z)=e^{z}$ and $f_{2}(w_{1})=iw_{1}$.  The composition of these maps is given by $g(z)=f_{2}(f_{1}(z))=ie^{z}$.  

These are just a few examples of how one can map a specific region to another region, which may be more useful for solving problems on, through a sequence of conformal maps.  The next interesting situation we may encounter is a region bounded by two non-intersecting circles, $C_{2}$ and $C_{1}$,  such that $C_{1}$ is in the interior of $C_{2}$.  We would prefer to work with circles that have the same center, so the challenge becomes mapping the region described above to an annulus.  

\begin{proposition} \label{centering circles} 
Let $C_{2}$ be the unit circle and let $C_{1}$ be a circle on the interior of $C_{2}$ centered on the $x$-axis with $x$-intercepts $(a,0)$ and $(b,0)$ with $|a|<b<1$.  Let $\Omega$ be the region bounded by $C_{1}$ and $C_{2}$.  Then, $\phi(z)=\frac{z-\alpha}{1-\alpha z}$ with $\alpha=\frac{1+ab}{a+b}-\sqrt{(\frac{1+ab}{a+b})^{2}-1}$ will map $C_{2}$ onto $C_{2}$, $C_{1}$ onto a circle centered at the origin with radius $r=\phi(b)$, and the region between $C_{2}$ and $C_{1}$ onto the annular region bounded $\phi[C_{1}]$ and the unit circle.  
\end{proposition} 
\begin{proof}
We will show that we can choose an $\alpha$ such that $-1<\alpha<1$ an $\phi_{\alpha}(z)=\frac{z-\alpha}{1-\overline{\alpha}z}$ maps the circle $C_{1}$ onto a circle centered at the origin.  It is known that this map maps the unit circle onto itself, so $C_{2}$ is mapped onto itself under $\phi_{\alpha}$ \cite{conway}.  It suffices to show choose an $\alpha$ so that $\phi_{\alpha}(a)=-\phi_{\alpha}(b)$.  The reason being is that we would like for $\phi_{\alpha}(a)$ and $\phi_{\alpha}(b)$ to serve as endpoints for the diameter of our new circle $\phi[C_{1}]$ centered at the origin.  This would imply that $\phi_{\alpha}(a)=-\phi_{\alpha}(b)$.  With this condition, we have
\begin{equation*}
\frac{a-\alpha}{1- \alpha a}=-\frac{b-\alpha}{1- \alpha b}.
\end{equation*}
Moreover, we have
\begin{equation*}
\alpha^2-2\frac{1+ab}{a+b}  \alpha +1=0.
\end{equation*}
It's important to note that $a+b\neq 0$ from the assumption that $|a|<b<1$.    
Using the quadratic equation to solve for $\alpha$, we obtain
\begin{equation*}
\alpha_1=\frac{1+ab}{a+b}+\sqrt{\left(\frac{1+ab}{a+b}\right)^2-1},  \, 
\alpha_2=\frac{1+ab}{a+b}-\sqrt{\left(\frac{1+ab}{a+b}\right)^2-1}. 
\end{equation*} 
We next proceed to show that $\alpha_1>1$ and $0<\alpha_2<1$, which means we must choose $\alpha_2$ in constructing $\phi_\alpha$.  To do so, we will first show that $1+ab \geq a+b$ given the assumptions that  $|a|<1$, $0<b <1$ and $0\leq |a|<b<1$.  Since $1-b>0$, if $0\leq a<1$, then
multiplying $1-b$ by $a$, we get $1-b> a (1-b)$.  If $-1<a<0$, then $1-b> a (1-b)$, because the right side is negative, while the left side is positive.  In all cases, $1+ab>a+b>0$.  Consequently, $\left(\frac{1+ab}{a+b}\right)^2>1$; hence the discriminant of the quadratic equation in $\alpha$ in part is positive and so we have two distinct roots, $\alpha_1$ and $\alpha_2$.  From the inequality $1+ab>a+b>0$, we conclude that 
$\frac{1+ab}{a+b}>1$ and so $\alpha_1>1$.  The product of $\alpha_1\cdot\alpha_2=1$, as can be checked directly or by using a well-known property that the product of the roots of 
the quadratic equation $ax^2+bx+c=0$ is $\frac{c}{a}$.  Since $1<\alpha_1$, we 
conclude that $0<\alpha_2<1$, in order to have the equality $\alpha_1\cdot\alpha_2=1$.  We conclude that $\phi(z)=\frac{z-\alpha}{1-\alpha z}$ with $\alpha=\frac{1+ab}{a+b}-\sqrt{\left(\frac{1+ab}{a+b}\right)^2-1}$
will map $C_2$ onto $C_2$, $C_1$ onto a circle centered at the origin with radius $r=\phi(b)$,
and the region between $C_2$ and $C_1$ onto the annular region bounded by $\phi[C_1]$ and the unit circle. 
\hfill\end{proof} 
Note: If $C_{2}$ is not the unit circle, we can find a mapping that shifts and scales $C_{2}$ by the appropriate increments so that we obtain the unit circle.  After doing so, we can apply the process established in Proposition \ref{centering circles}.
\begin{example} We derive the linear fractional transformation that maps the shaded region between the two given circles onto an annular region centered at the origin.  
\begin{figure}[H]
\centering  
\begin{minipage}[t]{0.48\textwidth}
\includegraphics[width=\textwidth]{annulus}
\caption{Here $a=-\frac14$, $b=\frac47$.}
\end{minipage} 
\end{figure}
\end{example}
\vspace{-0.15in}
\noindent \textbf{Solution.} 
Note that $a=-\frac{1}{4}$ and $b=\frac{4}{7}$ satisfies $|a|<b<1$, so we use the linear fractional transformation from Proposition \ref{centering circles}.  Using the formula for $\alpha$, we obtain $\alpha=\frac{8-\sqrt{55}}{3}$.  Thus, $\phi(z)=\frac{3z-(8-\sqrt{55})}{3-(8-\sqrt{55})z}$ from Proposition \ref{centering circles}.

\begin{example} \label{lftcircles}
We supply the formulas of the analytic mappings in each sequence of mappings shown in Figure \ref{lftcirclesfig} and derive the given formula for the final composite mapping $w=f(z)$.  
\begin{figure}[H]
\vspace{-2.85in}
\centering   \begin{minipage}[t]{0.99\textwidth}
\includegraphics[width=\textwidth]{ca6-2-28}\vspace{-3.15in}
\caption{Sequence of mappings for Example \ref{lftcircles}.} \label{lftcirclesfig}
\end{minipage} 
\vspace{-3.0in}
\end{figure}
\noindent \textbf{Solution.} The first map from the $z$-plane to the $w_{1}$-plane is given by $f_{1}(z)=\frac{1}{z}$.  The second map from the $w_{1}$-plane to the $w_{2}$-plane is simply a counterclockwise rotation by $\frac{\pi}{4}$ and is given by $f_{2}(w_{1})=(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})w_{1}$.  The last map from $w_{2}$-plane to the $w$-plane is of the type described in Proposition \ref{centering circles}.  Here $\alpha=\frac{1}{2}$ and so $f_{3}(w_{2})=\frac{2w_{2}-1}{2-w_{2}}$.  The composition $g(z)=f_{3}(f_{2}(f_{1}(z)))=\frac{2(\sqrt{2}+i\sqrt{2})-2z}{4z-(\sqrt{2}+i\sqrt{2})}$.
\end{example}

\chapter{The Schwarz-Christoffel Transformation} 
In this chapter, we describe another particular type of one-to-one, conformal mapping, the Schwarz-Christoffel transformation.  The Schwarz-Christoffel transformation will allow us to map the upper half-plane onto a polygonal region.  

Suppose $\Omega$ is a region in the $w$-plane, bounded by a positively oriented polygonal path $P$ with $n$ sides.  We will let $w_{1}, w_{2}, \ldots, w_{n}$ denote the vertices of $P$ counted as the path $P$ is traced with positive orientation.  Figure \ref{n=5} illustrates the case when $n=5$.  
If $P$ is bounded, then the point $w_{n}$ is taken to be the initial and terminal point of the closed path $P$.  If $P$ is unbounded, we specify $w_{n}=\infty$ and view $P$ as a polygon with $n-1$ vertices $w_{1}, w_{2}, \ldots, w_{n-1}$.  This is illustrated in Figure \ref{unbounded}.  Additionally, we let $\theta_{j}$ denote the angle created as we turn the corner of the polygon at $w_{j}$.  We choose $0<|\theta_{j}|<\pi$ for $j=1,\ldots,n$.  A positive value corresponds to a left turn; meanwhile, a negative value corresponds to a right turn.  We see in Figure \ref{n=5} that $\theta_{2}$ is the only negative angle.  
\vspace{-.4in}
\begin{figure}[H]
\hspace{.3in}
  \centering   \begin{minipage}[t]{0.33\textwidth}
    \includegraphics[width=\textwidth]{ca6-4-01}
   \caption{Positively oriented 
polygonal boundary with  
corner angles measured from the outside.}\label{n=5}
   \end{minipage} 
\hfill
  \centering  
   \begin{minipage}[t]{0.33\textwidth}
    \includegraphics[width=\textwidth]{ca6-4-02}
   \caption{Unbounded polygonal region with 4 sides and 3 vertices.}\label{unbounded}
   \end{minipage} 
\vspace{-.1in}
\hspace{.3in}
  \end{figure}

\begin{theorem}[Schwarz-Christoffel Transformation]\label{sctheorem}
Let $\Omega$ be a region bounded by a polygonal path $P$ with vertices at $w_{j}$ (counted consecutively) and corresponding exterior angles $\theta_{j}$.  Then there is a one-to-one conformal mapping $f$ of the upper half-plane onto $\Omega$, such that
\begin{equation}
f'(z)=A(z-x_{1})^{-\frac{\theta_{1}}{\pi}}(z-x_{2})^{-\frac{\theta_{2}}{\pi}}\cdots(z-x_{n-1})^{-\frac{\theta_{n-1}}{\pi}},\label{grosseq}
\end{equation}
where $A$ is a constant, the $x_{j}'s$ are real and satisfy $x_{1}<x_{2}<\cdots<x_{n-1}$, $f(x_{j})=w_{j}$, $\lim_{z \rightarrow \infty} f(z)=w_{n}$, and the complex powers are defined by their principal branches. 
\end{theorem}
The points $x_{j}$ on the $x$-axis are the pre-images of the vertices of the polygonal path $P$ in the $w$-plane.  We may choose two of the $x_{j}$'s arbitrarily as long as they are arranged in ascending order.  Similar to our discussion in the previous chapter we denote $\lim_{z \rightarrow \infty} f(z)=w_{n}$ as $f(\infty)=w_{n}$.  If $P$ is unbounded, we have $f(\infty)=\infty$.  
\begin{definition}
The mapping $f$ whose derivative is given in Equation \ref{grosseq} is called a \textit{Schwarz-Christoffel transformation}, which is named after the German mathematicians Karl Herman Amandus Schwarz (1843-1921) and Elwin Bruno Christoffel (1829-1900).  Since $f$ is an antiderivative of Equation \ref{grosseq}, we can write 
\begin{equation}
f(z)=A \int (z-x_{1})^{-\frac{\theta_{1}}{\pi}}(z-x_{2})^{-\frac{\theta_{2}}{\pi}} \cdots (z-x_{n-1})^{-\frac{\theta_{n-1}}{\pi}} \, dz+B.  \label{grossereq}
\end{equation} 
Here the constants $A$ and $B$ depend on the polygonal path $P$.  
\end{definition}
Proving Theorem \ref{sctheorem} is quite difficult, so we provide a sketch of the proof and a proof in the case of the upper half-plane being mapped onto an arbitrary triangle.  The sketch of the general proof for a closed polygon requires the following lemma, which we prove first.  
\begin{lemma}[Closure of the polygon] \label{closurelemma}
Let $P$ be a closed polygon with  $\theta_{1}+\cdots+\theta_{n}=2\pi$ and $f$ be given as in Equation \ref{grossereq}.  Then, $f(z) \rightarrow w_{n}$ as $z \rightarrow \infty$.  
\end{lemma}
\begin{proof}
Suppose $\theta_{1}+\theta_{2}+\cdots+\theta_{n}=2\pi$.  We can safely assume $\theta_{n}<\pi$ by the assumption that we can choose $0<|\theta_{j}|<\pi$ for $j=1,\ldots,n$.  It follows that
\begin{equation*}
2\pi=\theta_{1}+\theta_{2}+\cdots+\theta_{n}<\theta_{1}+\theta_{2}+\cdots+\theta_{n-1}+\pi,
\end{equation*}
so we conclude that $\pi<\theta_{1}+\theta_{2}+\cdots+\theta_{n-1}$ by subtracting $\pi$ from both sides of the inequality above.  We conclude that $\theta_{1}+\theta_{2}+\cdots+\theta_{n-1}>\pi$.  We let $\beta_{j}=\frac{\theta_{j}}{\pi}$.


Because $A$ and $B$ in Equation \ref{grosseq} just dilate, rotate, and translate the mapping, $A$ and $B$ do not affect the convergence of $f$ or the closure of the polygon.  Hence, we can choose $A=1$ and $B=0$.  Let $x_{0}=1+\max_{1 \leq j \leq n-1}x_{j}$ and set 
\begin{equation*}
f(z)=\int_{[x_{0},z]}\frac{d\zeta}{(\zeta-x_{1})^{\beta_{1}}(\zeta-x_{2})^{\beta_{2}}\cdots(\zeta-x_{n-1})^{\beta_{n-1}}}.
\end{equation*}


Next, let $x_{0}=1+\text{max}_{1 \leq j \leq n-1}x_{j}$ and $z=x$ be real.  We will use the limit comparison test for the integrand to show that 
\begin{equation*}
\lim_{y \rightarrow \infty} f(y)=\int_{x_{0}}^{\infty}\frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}}
\end{equation*}
is finite.  Let $a=\text{max}_{1 \leq j \leq n-1} x_{j}$ and $b=\text{min}_{1 \leq j \leq n-1} x_{j}$.  Observe that
\begin{align*}
x-a+a-b&=(x-a)\left(1+\frac{a-b}{x-a}\right) \\
&\leq (x-a)(1+a-b)
\end{align*}
because $x-a \geq 1$.  This observation gives us 
\begin{equation*}
(1+a-b)(x-a) \geq x-a+a-b=x-b \geq x-x_{j} \geq x-a
\end{equation*}
for all $1 \leq j \leq n-1$.  We see that $(x-x_{j})^{\beta_{j}} \approx (x-a)^{\beta_{j}}$ for all $\beta_{j}$.  Using this, we have 
\begin{equation*}
\int_{x_{0}}^{\infty} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}} \leq C \int_{x_{0}}^{\infty} \frac{dx}{(x-a)^{\beta_{1}+\beta_{2}+\cdots+\beta_{n-1}}},
\end{equation*}
where $C$ is a constant.  Specifically, we have
\begin{equation*}
C=\prod_{j: \beta_{j}>0} (1+a-b)^{\beta_{j}}.
\end{equation*}
Notice that 
\begin{equation*}
\beta_{1}+\beta_{2}+\cdots+\beta_{n-1}=\frac{\theta_{1}+\theta_{2}+\cdots+\theta_{n-1}}{\pi}>\frac{\pi}{\pi}=1
\end{equation*}
from our previous work above.  Therefore, after a change in variables, we see that the integral on the right converges.  


Lastly, let 
\begin{equation*}
w_{n}=\lim_{y \rightarrow \infty} f(y)=\int_{x_{0}}^{\infty} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}}
\end{equation*}
from above.  We want to show that $\lim_{|z| \rightarrow \infty}f(z)=w_{n}$ for any $z \in \mathbb{C}$.  We will prove this for $z=Re^{i\theta}$ where $R>0$ and $0 \leq \theta \leq \pi$.  A similar argument will hold for $z=Re^{i\theta}$ where $R>0$ and $\pi \leq \theta \leq 2\pi$.  First, observe that 
\begin{align*}
|f(Re^{i\theta})-f(R)|&=\bigg|\int_{[x_{0},Re^{i\theta}]} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}} \\
& \hspace{1cm}-\int_{[x_{0},R]} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}}\bigg| \\
&=\bigg|\int_{[x_{0},Re^{i\theta}]} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}} \\
& \hspace{1cm}+\int_{[R,x_{0}]} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}}\bigg|\\
&=\left|\int_{[R,Re^{i\theta}]} \frac{dx}{(x-x_{1})^{\beta_{1}}(x-x_{2})^{\beta_{2}}\cdots(x-x_{n-1})^{\beta_{n-1}}}\right| \\
&\leq R\pi M(R), 
\end{align*}
where $M(R)$ is the maximum of the absolute value of the integrand on the upper semicircle of radius $R$.  
Also, the length of the segment joining $R$ to $Re^{i\theta}$ is largest when it is the circumference of the upper semicircle or $R\pi$.  Next, we claim that $\lim_{R \rightarrow \infty} M(R)R^{\beta_{1}+\cdots+\beta_{n-1}}=1$.  Notice that 
for $R>\max_{1\le j \le n-1}  |x_j|  $ we have
\begin{align*}
M(R)&=\sup_{|z|=R}\left|\frac{1}{(z-x_{1})^{\beta_{1}}\cdots(z-x_{n-1})^{\beta_{n-1}}}\right| \\
&=\sup_{|z|=R}\left[ \prod_{\substack{1\le j \le n-1\\ \beta_j \ge 0}} \frac{1}{|z-x_{j}|^{\beta_{j}}} 
\prod_{\substack{1\le j \le n-1 \\ \beta_j<0}} \frac{1}{|z-x_{j}|^{\beta_{j}}} \right] \\
&= \sup_{|z|=R}\left[ \prod_{\substack{1\le j \le n-1\\ \beta_j \ge 0}} \frac{1}{|z-x_{j}|^{| \beta_{j}| }} 
\prod_{\substack{1\le j \le n-1 \\ \beta_j<0}}  |z-x_{j}|^{|\beta_{j}|} \right] \\
&=    \prod_{\substack{1\le j \le n-1\\ \beta_j \ge 0}} \frac{1}{(R-x_{j})^{| \beta_{j}| }} 
\prod_{\substack{1\le j \le n-1 \\ \beta_j<0}}  (R+x_{j})^{|\beta_{j}|} ,  
\end{align*} 
since the first product becomes maximum for $z=R$ and the second product becomes maximum for $z=-R$ and both of these numbers lie on the circle $|z|=R$. 
This implies that 
\begin{equation*}
\lim_{R \rightarrow \infty} M(R)R^{\beta_{1}+\cdots+\beta_{n-1}}=\lim_{R \rightarrow \infty} 
 \prod_{\substack{1\le j \le n-1\\ \beta_j \ge 0}} \frac{R^ {| \beta_{j}| } }{(R-x_{j})^{| \beta_{j}| }} 
\prod_{\substack{1\le j \le n-1 \\ \beta_j<0}}  \frac{ (R+x_{j})^{|\beta_{j}|} }{R^ {| \beta_{j}| }}.
\end{equation*}
But, we have $\beta_{i} \geq 0$ 
\begin{equation*}
 \lim_{R \rightarrow \infty} \frac{R^{\beta_{i}}}{(R-x_{i})^{\beta_{i}}}=1 
\end{equation*}
and for  $\beta_{j}<0$, 
\begin{equation*}
 \lim_{R \rightarrow \infty} \frac{(x_{j}+R)^{|\beta_{j}|}}{R^{|\beta_{j}|}}=1.
\end{equation*}
We conclude that $\lim_{R \rightarrow \infty}M(R)=1$.  Note that 
\begin{equation*}
RM(R)=R^{1-(\beta_{1}+\cdots+\beta_{n-1})}(M(R)R^{\beta_{1}+\cdots+\beta_{n-1}}).
\end{equation*}
The fact that $\beta_{1}+\cdots+\beta_{n-1}>1$ by our discussion in part (c) implies that $RM(R) \rightarrow 0$ as $R \rightarrow \infty$.  This implies that $|f(Re^{i \theta})-f(R)| \rightarrow 0$ uniformly in $\theta$ as $R \rightarrow \infty$.  The fact that $f(R) \rightarrow w_{n}$ as $R \rightarrow \infty$ implies that $f(z) \rightarrow w_{n}$ as $|z|\rightarrow \infty$.  
\hfill\end{proof}
\begin{proof}
Let $f$ be such a function whose derivative is given by Equation \ref{grosseq}.  Let $w_{j}=f(x_{j})$ for $1<j<n-1$ with $x_{1}<x_{2}<\cdots<x_{n-1}$.  For any point $x_{0} \in \mathbb{R}$, the slope of the line segment containing $f(x_{0})$ is given by $\text{arg}(f'(x_{0}))$ by Theorem \ref{conformal theorem}.  By the properties of the argument, we have 
\begin{equation*}
\text{arg}(f'(z))=\text{arg}(A)-\frac{\theta_{1}}{\pi}\text{arg}(z-x_{1})-\frac{\theta_{2}}{\pi}\text{arg}(z-x_{2})-\cdots-\frac{\theta_{n-1}}{\pi}\text{arg}(z-x_{n-1}).
\end{equation*}
For $z \in \mathbb{R}$ with $z<x_{1}$, we will have $z<x_{j}$ for all $1<j<n-1$.  This implies that $z-x_{j}<0$ for all $1<j<n-1$, which means $\text{arg}(z-x_{j})=\pi$ for all $1<j<n-1$.  Thus, $\text{arg}(f'(z))=\text{arg}(A)-(\theta_{1}+\theta_{2}+\cdots+\theta_{n-1})$.  If $w_{n}=f(\infty)$, all the points in the interval $(-\infty,x_{1})$ are mapped onto a line segment starting with $w_{n}$ and ending with $w_{1}=f(x_{1})$ and at an angle given by $\text{arg}(f'(z))=\text{arg}(A)-(\theta_{1}+\theta_{2}+\cdots+\theta_{n-1})$.  For the points in the interval $(x_{1},x_{2})$, we have $z-x_{1}>0$ and $z-x_{j}<0$ for all $2<j<n-1$.  Then, $\text{arg}(z-x_{1})=0$ and $\text{arg}(z-x_{j})=\pi$ for $2<j<n-1$.  Here we have $\text{arg}(f'(z))=\text{arg}(A)-(\theta_{2}+\cdots+\theta_{n-1})$.  At the point $x_{1}$, the path turns by an angle of $\theta_{1}$ and all the points in $(x_{1},x_{2})$ are mapped onto the line segment with initial point $w_{1}$ and terminal point $w_{2}=f(x_{2})$.  We continue this process until we have the points in the interval $(x_{n-1},\infty)$ being mapped onto a line segment with initial point $w_{n-1}$ and at angle $\text{arg}(f'(z))=\text{arg}(A)$.  If we have bounded polygon, this line segment will connect back to $w_{n}$ by Lemma \ref{closurelemma}.  The resulting polygon is closed and $\theta_{1}+\cdots+\theta_{n}=2\pi$.  This shows that $f$ takes the real line onto a polygon with vertices $w_{j}=f(x_{j})$ and exterior angles $\theta_{j}$.  Since the upper half-plane is on our left as we traverse the real line, we know that interior our the resulting polygon $P$ will be the left as we traverse the polygon with positive orientation.  
\hfill\end{proof}
\begin{proof}  We prove Theorem \ref{sctheorem} in the case where the upper half-plane is mapped to an arbitrary triangle.  The argument we provide is based on geometric considerations and can be made to work for general polygons.  
\begin{figure}[H]
  \centering   \begin{minipage}[t]{1.0\textwidth}
    \includegraphics[width=\textwidth]{halfplane}
   \caption{Here $\Omega$ is the upper half-plane.}
   \end{minipage} 
   \end{figure}
   \begin{figure}[H] 
   \vspace{2.5in}
  \centering   \begin{minipage}[t]{1.0\textwidth}
 \vspace{-2.5in}   \includegraphics[width=\textwidth]{triangle}
   \caption{The image of $\Omega$ under a Schwarz-Christoffel transformation $f$ is the enclosed triangle.} \label{triangle}
   \end{minipage} 
  \end{figure}
Notice that $\text{arg}(f'(y_{2})),\text{arg}(f'(y_{3}))>0$; meanwhile, $\text{arg}(f'(y_{1}))<0$.  Let 
\begin{align*}
\varphi_{1}&=\text{arg}(f'(y_{1}))+\pi \\ 
\varphi_{2}&=\text{arg}(f'(y_{2})) \\
\varphi_{3}&=\text{arg}(f'(y_{3})) 
\end{align*}
to ensure that $\varphi_{j}>0$ for all $j=1,2,3$.  We want to verify the following
\begin{enumerate}[(1)]
\item $\text{arg}(f'(y_{1}))=\text{arg}(A)-(\theta_{1}+\theta_{2})$ 
\item $\text{arg}(f'(y_{2}))=\text{arg}(A)-\theta_{2}$
\end{enumerate}
with the assumption that $\text{arg}(f'(y_{3}))=\text{arg}(A)$.  For (2), notice that $\varphi_{3}=\varphi_{2}+\theta_{2}$ by the fact that the an exterior angle is the sum of the two interior angles.  Thus, we have established (2).  For (1), we will first find $\omega$.  Here 
\begin{equation*}
\omega=\pi-(\pi-\theta_{1}+\pi-\theta_{2})=\theta_{1}+\theta_{2}-\pi.
\end{equation*}
This implies $\varphi_{3}=\varphi_{1}+(\theta_{1}+\theta_{2}-\pi)$, which establishes (1).  
\hfill\end{proof}
\begin{example} \label{sc1} We find the Schwarz-Christoffel transformation of the upper half-plane onto the region given by Figure \ref{1part(a)} and Figure \ref{1part(b)}. 
\begin{figure}[H]
\vspace{-.4in}
   \begin{minipage}[t]{0.32\textwidth}
    \includegraphics[width=\textwidth]{ca6-4-18}\vspace{-.2in}
   \caption{Part (a).} \label{1part(a)}
   \end{minipage} 
   \hspace{0.75in}
  \centering   \begin{minipage}[t]{0.32\textwidth}
    \includegraphics[width=\textwidth]{ca6-4-19}\vspace{-.2in}
   \caption{Part (b).} \label{1part(b)}
   \end{minipage} 
\vspace{-.3in}
  \end{figure}
\end{example}
\noindent \textbf{Solution}
\begin{enumerate}
\item[(a)]
We have that $x_{1}=-1$, $x_{2}=1$, $\theta_{1}=-\pi$, and $\theta_{2}=\frac{\pi}{2}$.  Using Equation \ref{grossereq}, we obtain 
\begin{align*}
f(z)&=A\int (z+1)(z-1)^{-\frac{1}{2}} \, dz+B \\ 
&=A \int \frac{z+1}{(z-1)^{\frac{1}{2}}} \, dz + B \\ 
&=A \int (z-1)^{\frac{1}{2}}+\frac{2}{(z-1)^{\frac{1}{2}}} \, dz +B  \\
&=A\left[\frac{2}{3} (z-1)^{\frac{3}{2}}+4(z-1)^{\frac{1}{2}}\right]+B.
\end{align*}
The fact that $f(1)=1$ implies $B=1$.  The fact that $f(-1)=0$ implies that $0=A \left(\dfrac{2^{\frac{7}{2}}i}{3}\right)+1$.  Thus, $A=\frac{3i}{2^{\frac{7}{2}}}$.  We conclude that 
\begin{equation*}
f(z)=\frac{3i}{2^{\frac{7}{2}}}\left[\frac{2}{3}(z-1)^{\frac{3}{2}}+4(z-1)^{\frac{1}{2}}\right]+1.
\end{equation*}
\item[(b)] We have that $x_{1}=-1$, $x_{2}=1$, $\theta_{1}=-\frac{\pi}{2}$, and $\theta_{2}=\frac{\pi}{2}$.  Using Equation \ref{grossereq}, we have 
\begin{align*}
f(z)&=A \int (z+1)^{\frac{1}{2}}(z-1)^{-\frac{1}{2}} \, dz+B  \\
&=A \int \frac{(z+1)^{\frac{1}{2}}}{(z-1)^{\frac{1}{2}}} \, dz +B \\
&=A \int \frac{z}{i(1-z^{2})^{\frac{1}{2}}}+\frac{1}{i(1-z^{2})^{\frac{1}{2}}} \, dz+B  \\
&=A[i(1-z^{2})^{\frac{1}{2}}-i\sin^{-1}(z)]+B \\ 
&=Ai[(1-z^{2})^{\frac{1}{2}}-\sin^{-1}(z)]+B.
\end{align*}
The fact that $f(-1)=0$ implies that $\frac{A\pi i}{2}+B=0$.  The fact that $f(1)=-i$ implies that $-\frac{A \pi i}{2}+B=-i$.  Solving this system of equations yields $B=-\frac{i}{2}$ and $A=\frac{1}{\pi}$.  Thus, we have 
\begin{equation*}
f(z)=\frac{i}{\pi}[(1-z^{2})^{\frac{1}{2}}-\sin^{-1}(z)]-\frac{i}{2}. 
\end{equation*}
\end{enumerate}
\begin{example} We find the Schwarz-Christoffel transformation of the upper half-plane onto the region given by Figure \ref{2part(a)} and Figure \ref{2part(b)}.
\begin{figure}[H]
\vspace{-.3in}
  \centering   \begin{minipage}[t]{0.32\textwidth}
    \includegraphics[width=\textwidth]{ca6-4-20}\vspace{-.2in}
   \caption{Part (a).}\label{2part(a)}
   \end{minipage} 
   \hspace{0.75in}
  \centering   \begin{minipage}[t]{0.32\textwidth}
 \vspace{-2.5in}   \includegraphics[width=\textwidth]{ca6-4-22}\vspace{-.2in}
   \caption{Part (b).}\label{2part(b)}
   \end{minipage} 
\vspace{-.1in}
  \end{figure}
\end{example}
\noindent \textbf{Solution} 
\begin{enumerate}
\item[(a)] 
We have that $x_{1}=-1$, $x_{2}=1$, $\theta_{1}=\frac{\pi}{2}$, and $\theta_{2}=-\frac{\pi}{2}$.  Using Equation \ref{grossereq}, we have 
\begin{align*}
f(z)&=A \int (z+1)^{-\frac{1}{2}}(z-1)^{\frac{1}{2}} \, dz +B \\ 
&=A \int \frac{(z-1)^{\frac{1}{2}}}{(z+1)^{\frac{1}{2}}} \, dz+B \\ 
&=A \int \frac{z}{i(1-z^{2})^{\frac{1}{2}}}-\frac{1}{i(1-z^{2})^{\frac{1}{2}}} \, dz +B \\ 
&=A[i(1-z^{2})^{\frac{1}{2}}+i\sin^{-1}(z)]+B \\ 
&=Ai[(1-z^{2})^{\frac{1}{2}}+\sin^{-1}(z)]+B. 
\end{align*}
The fact that $f(-1)=0$ implies that $-\frac{A \pi i}{2}+B=0$.  The fact that $f(1)=i$ implies that $\frac{A \pi i}{2}+B=i$.  Solving the system of equations yields $B=\frac{i}{2}$ and $A=\frac{1}{\pi}$.  Thus, we have that 
\begin{equation*}
f(z)=\frac{i}{\pi}\left[(1-z^{2})^{\frac{1}{2}}+\sin^{-1}(z)\right]+\frac{i}{2}. 
\end{equation*}
\item[(b)] We have that $x_{1}=-1$, $x_{2}=1$, $\theta_{1}=\pi$, and $\theta_{2}=-\frac{\pi}{2}$.  Using Equation \ref{grossereq}, we have
\begin{align*}
f(z)&=A \int (z+1)^{-1}(z-1)^{\frac{1}{2}} \, dz+B \\
&=A \int \frac{1}{\sqrt{z-1}}-\frac{2}{(\sqrt{z-1})(z+1)} \, dz +B  \\
&=A \int \frac{1}{\sqrt{z-1}}-\frac{i}{\sqrt{2}}\left[\frac{1}{\sqrt{z-1}}\left(\frac{1}{\sqrt{z-1}+i\sqrt{2}}-\frac{1}{\sqrt{z-1}-i\sqrt{2}}\right)\right] \,dz +B \\
&=2A\left(\sqrt{z-1}-\frac{i}{\sqrt{2}}\text{Log}(\sqrt{z-1}+i\sqrt{2})+\frac{i}{\sqrt{2}}\text{Log}(\sqrt{z-1}-i\sqrt{2})\right)+B.
\end{align*}
The fact that $f(1)=0$ implies $2A\left(\frac{\pi}{\sqrt{2}}\right)+B=0$.  First, we will find $A$.  Notice that $\text{Arg}(A)=-\frac{\pi}{2}$ as that is the measure of the exterior angle of the final line segment.  We conclude that $A=-i|A|$.  Substituting this into the above equation yields $B=\sqrt{2}i|A|\pi$.  Finding $A$ and $B$ now reduces to finding $|A|$.  Notice that $\text{Im}(f(x))=1$ for $x<-1$.  Consider $x=-2$ and notice that 
\begin{equation*}
f(-2)=-2|A|\left(-\sqrt{3}+\frac{1}{\sqrt{2}}\ln|(\sqrt{3}+\sqrt{2})i|-\frac{1}{\sqrt{2}}\ln|(\sqrt{3}-\sqrt{2})i|\right)+\sqrt{2}i|A|\pi,
\end{equation*}
so $\text{Im}(f(-2))=\sqrt{2}|A|\pi$.  But, $\text{Im}(f(-2))=1$.  We have that $\sqrt{2}|A|\pi=1$ and so $|A|=\frac{1}{\sqrt{2}\pi}$.  Thus, $A=\frac{-i}{\sqrt{2}\pi}=\frac{-i\sqrt{2}}{2\pi}$ and $B=i$.  Substituting this into $f$ yields
\begin{equation*}
f(z)=\frac{-i\sqrt{2}}{\pi}\left(\sqrt{z-1}-\frac{i}{\sqrt{2}}\text{Log}(\sqrt{z-1}+i\sqrt{2})+\frac{i}{\sqrt{2}}\text{Log}(\sqrt{z-1}-i\sqrt{2})\right)+i.
\end{equation*}
\end{enumerate}
\begin{example}[Channel width formula] 
Observe that we can write the logarithm as a Schwarz-Christoffel transformation by letting $\text{Log}(z)=\int \frac{dz}{z}$ where there is only one vertex at $x_{1}=0$ having $\theta_{1}=\pi$ and $w_{1}=\infty$.  Choosing  $\theta_{1}=\pi$ forces a semi-infinite channel in the left half-plane.  Here we will also have an infinite channel in the right half-plane, as shown in Figure \ref{channel}. 

 \begin{figure}[H] 
   \vspace{2.5in}
  \centering   \begin{minipage}[t]{0.4\textwidth}
 \vspace{-2.5in}   \includegraphics[width=\textwidth]{ca6-4-23}\vspace{-.6in}
   \caption{The channel width $|s|$.} \label{channel}
   \end{minipage} 
\end{figure} 

\noindent For points $x^{*}$ on the real axis near $x_{1}=0$, we have 
\begin{equation*}
f(x^{*})-f(-x^{*})=-i\pi,
\end{equation*}
so the channel width is $\pi$.


First, we consider a Schwarz-Christoffel transformation where $\theta_{j_{0}}=\pi$ and $\theta_{j} \leq \pi$ for all $j \neq j_{0}$.  By Equation \ref{grossereq}, we have 
\begin{equation*}
f(z)=A \int \frac{dz}{(z-x_{1})^{\frac{\theta_{1}}{\pi}}\cdots(z-x_{j_{0}})\cdots(z-x_{n-1})^{\frac{\theta_{n-1}}{\pi}}}+B.
\end{equation*}
We will now that show that $\lim_{z \rightarrow x_{j_{0}}} |f(z)|=\infty$ as $z \rightarrow z_{j_{0}}$ from the upper half-plane.  Notice that 
\begin{equation*}
f'(z)=\frac{1}{(z-x_{1})^{\frac{\theta_{1}}{\pi}}\cdots(z-x_{j_{0}})\cdots(z-x_{n-1})^{\frac{\theta_{n-1}}{\pi}}}.
\end{equation*}
Let $f'(z)=\frac{g(z)}{z-x_{j_{0}}}$ where $g(z)=(z-x_{1})^{-\frac{\theta_{1}}{\pi}}\cdots(z-x_{n-1})^{-\frac{\theta_{n-1}}{\pi}}$.  Here $g(z)$ is analytic near $x_{j_{0}}$ and has a Taylor Series expansion about $x_{j_{0}}$.  Then, 
\begin{equation*}
f'(z)=\frac{a_{0}}{z-x_{j_{0}}}+a_{1}+a_{2}(z-x_{j_{0}})+a_{3}(z-x_{j_{0}})^{2}+\cdots.
\end{equation*}
This implies that $f(z)=\text{Log}_{-\frac{\pi}{2}}(z-x_{j_{0}})+h(z)$ where $h(z)$ is analytic.  Moreover, $f(z)=\ln|z-x_{j_{0}}|+i\text{Arg}_{-\frac{\pi}{2}}(z-x_{j_{0}})+h(z)$.  We see that $\lim_{z \rightarrow x_{j_{0}}}|f(z)|=\infty$ because $\lim_{z \rightarrow x_{j_{0}}} \ln|z-x_{j_{0}}|=\infty$. 


Next, we'll define the channel seperation complex number $s$ by 
\begin{equation*}
s=\lim_{r \downarrow 0} (f(x_{j_{0}}+r)-f(x_{j_{0}}-r)).
\end{equation*}
We see $s$ labeled in Figure \ref{channel} and $|s|$ will give us the channel width.  Observe that we can parameterize the upper-half semi-circle from $x_{j_{0}}-r$ to $x_{j_{0}}+r$ as $z(t)=x_{j_{0}}+re^{i(\pi-t)}$ for $0 \leq t \leq \pi$.  Suppose that $s=\lim_{r \downarrow 0} (f(x_{j_{0}}+r)-f(x_{j_{0}}-r))$.  Notice that
\begin{align*}
s&=\lim_{r \downarrow 0} [f(x_{j_{0}}+r)-f(x_{j_{0}}-r)] \\
&=\lim_{r \downarrow 0} [f(z(\pi))-f(z(0))] \\
&=\lim_{r \downarrow 0} f(z(t))\big|_{0}^{\pi} \\
&=\lim_{r \downarrow 0} \int_{0}^{\pi} f'(z(t))z'(t) \, dt \\ 
&=A\lim_{r \downarrow 0} \int_{0}^{\pi} \frac{z'(t) \, dt}{(z(t)-x_{1})^{\frac{\theta_{1}}{\pi}}\cdots(z(t)-x_{j_{0}})\cdots(z(t)-x_{n-1})^{\frac{\theta_{n-1}}{\pi}}} \\
&=A \lim_{r \downarrow 0} \int_{0}^{\pi} \frac{-ire^{i(\pi-t)}}{(x_{j_{0}}+re^{i(\pi-t)}-x_{1})^{\frac{\theta_{1}}{\pi}}\cdots(re^{i(\pi-t)})\cdots(x_{j_{0}}+re^{i(\pi-t)}-x_{n-1})^{\frac{\theta_{n-1}}{\pi}}} \\
&=A\int_{0}^{\pi} \lim_{r \downarrow 0} \frac{-i}{(x_{j_{0}}+re^{i(\pi-t)}-x_{1})^{\frac{\theta_{1}}{\pi}}\cdots(x_{j_{0}}+re^{i(\pi-t)}-x_{n-1})^{\frac{\theta_{n-1}}{\pi}}} \\
&=A\int_{0}^{\pi} \frac{-i}{\prod_{j=1, j \neq j_{0}}^{n-1}(x_{j_{0}}-x_{j})^{\frac{\theta_{j}}{\pi}}} \\ 
&=\frac{-A\pi}{\prod_{j=1, j \neq j_{0}}^{n-1}(x_{j_{0}}-x_{j})^{\frac{\theta_{j}}{\pi}}}.
\end{align*} 
Therefore, we see that
\begin{equation*}
|s|=\frac{|A|\pi}{\prod_{j=1, j \neq j_{0}}^{n-1}|x_{j_{0}}-x_{j}|^{\frac{\theta_{j}}{\pi}}}.
\end{equation*}
\end{example}

We conclude with a brief remark of the importance of these results.  
When studying fluid flow, we may be interested in constructing the flow of an ideal fluid within a polygonal domain $P$.  It turns out that a Schwarz-Christoffel transformation will be useful in solving this problem, which is outlined in Section 7.3-Section 7.5 in \cite{zill}.  
The Schwarz-Christoffel transformation can also be instrumental in the determination of streamlines and isotherms.  Several examples can be found in Section 6.4 in \cite{asmar}.  One can also use a Schwarz-Christoffel transformation in the calculation of air gap reluctance and inductance for magnetic circuits \cite{airgap}.  Furthermore, researchers have investigated using these particular transformations to produce grid generation algorithms for global ocean general circulation models \cite{oceans}.  We see that these transformations are used mostly in physics and engineering, but could be useful in other fields as well.  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\addcontentsline{toc}{chapter}{Bibliography}
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\bibitem{AsmarGrafakos} N. Asmar and L. Grafakos, \textit{Complex Analysis with Applications}, Undergraduate Texts in Mathematics, Springer, New York, NY, 2018, to appear. 

\bibitem{airgap} A. Balakrishnan, W. Joines, and T. Wilson, \textit{Air gap reluctance and inductance calculations for magnetic circuits using a shwarz-christoffel transformation}, IEEE Transactions on Power Electronics \textbf{12} (1997), 654-663. 

\bibitem{book1} R. Churchill and J. Brown, \textit{Complex Variables and Applications}, International Series in Pure and Applied Mathematics, McGraw-Hill, New York, NY, 1990.

\bibitem{conway} J. Conway, \textit{Functions of One Complex Variable}, Graduate Texts in Mathematics: \textbf{11}, Springer-Verlag, New York, NY, 1973.  

\bibitem{book2} J. Taylor, \textit{Complex Variables}, Pure and Applied Undergraduate Texts: \textbf{16}, American Mathematical Society, Providence, RI, 2011.  

\bibitem{oceans} B. Wang and B. Liu, \textit{On the use of schwarz-christoffel conformal mappings to the grid generation for global ocean models}, Geoscientific Model Development \textbf{8} (2015), 3471-3485.

\bibitem{zill} D. Zill and P. Shanahan, \textit{Complex Analysis}, Jones \& Bartlett Learning LLC, Burlington, MA, 2013.  

\end{thebibliography}
\end{document}

