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\begin{document}
\begin{abstract}
This is a list of solved problems on hyperbolic geometry, complementing the problems in Problem Sets 4 and 5 and the lectures and discussions on the topic.\\
\end{abstract}

\title{MAT 141: Solutions to Problems on hyperbolic geometry}

\maketitle

\vspace{-1cm}

\section{Problems on hyperbolic distances}

{\bf Problem 1}. For each of the following pairs of points $P,Q\in\H^2$, compute the hyperbolic distance $d_{\H^2}(P,Q)$:

\begin{enumerate}
    \item $P=3i$ and $Q=6i$.
    \item $P=2i+7$ and $Q=5i+7$.
    \item $P=i$ and $Q=1+2i$.
    \item $P=i$ and $Q=\rho+i$ where $\rho\in\R$ is any given real number.\\
\end{enumerate}
\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item We've seen in class that the hyperbolic distance between two points $(x,y_1),(x,y_2)$ with $y_1<y_2$ is $\ln\left(\frac{y_2}{y_1}\right)$. Thus, $d_{\H^2}(P,Q)=\ln(2)$.
    \item Using the formula above, $d_{\H^2}(P,Q)=\ln\left(\frac52\right)$.
    \item We first construct the geodesic through $P,Q$, then compute the distance along the geodesic segment from $P$ to $Q$. The line segment $\overline{PQ}$ has slope $1$, so its perpendicular bisector has slope $-1$. The perpendicular bisector passes through the midpoint $\frac{P+Q}{2}=\frac12+\frac32i$, so the equation of the perpendicular bisector is $(y-\frac32)=-(x-\frac12)$. When the perpendicular bisector intersects the $x$-axis, we have $y=0$, and hence $-\frac32=-x+\frac12\implies x=2$, so the center of the geodesic is $(2,0)$. The radius is the Euclidean distance from $(2,0)$ to $P$ or $Q$, so we compute $d_{\mathbb{E}}((2,0),(0,1))=\sqrt{5}$, and thus the geodesic has equation $(x-2)^2+y^2=5$. To find the distance from $P$ to $Q$, we use the formula $ds=\frac{\sqrt{1+(dy/dx)^2}}{y}dx$. We rewrite the geodesic equation to get $y=\sqrt{5-(x-2)^2}$, and $\frac{dy}{dx}=\frac{-(x-2)}{\sqrt{5-(2-x)^2}}$. From $P$ to $Q$, we have $0\leq x\leq 1$, thus 
    \[
    d_{\H^2}(P,Q)=\int_0^1\frac{\sqrt{1+\frac{(-(x-2))^2}{5-(x-2)^2}}}{\sqrt{5-(x-2)^2}}dx=\ldots=\int_0^1\frac{\sqrt{5}}{5-(2-x)^2}dx.
    \]
    You do not need to simplify further.
    \item We find the equation of the corresponding geodesic: The center is $\frac{\rho}{2}$, and the radius is $d_\mathbb{E}((\frac{\rho}{2},0),(0,1))=\sqrt{\frac{\rho^2}{4}+1}=\sqrt{\frac{\rho^2+4}{4}}$, thus the geodesic is $(x-\frac{\rho}{2})^2+y^2=\frac{\rho^2+4}{4}$. We parametrize this by $(x(t),y(t))=\left(\frac{\rho}{2}+\sqrt{\frac{\rho^2+4}{4}}\cos t,\sqrt{\frac{\rho^2+4}{4}}\sin t\right)$. For a general parametrization $(x(t),y(t))=(\alpha+r\cos t,r\sin t)$, we have the formula $ds=\frac{\sqrt{(dx/dt)^2+(dy/dt)^2}}{y(t)}dt=\frac{\sqrt{r^2\sin^2 t+r^2\cos^2 t}}{r\sin t}dt=\frac{1}{\sin t}dt$. Now, for $\rho>0$, we have the lower bound for $t$ satisfies $\frac{\rho}{2}+\sqrt{\frac{\rho^2+4}{4}}\cos t=\rho\implies t=\cos^{-1}\left(\frac{\rho}{2}\cdot\sqrt{\frac{4}{\rho^2+4}}\right)=\cos^{-1}\left(\frac{\rho}{\sqrt{\rho^2+4}}\right)=:t_1$, and the upper bound satisfies $\frac{\rho}{2}+\sqrt{\frac{\rho^2+4}{4}}\cos t=0\implies t=\cos^{-1}\left(\frac{-\rho}{\sqrt{\rho^2+4}}\right)=:t_2$. Thus, $d_{\H^2}(P,Q)=\int_{t_1}^{t_2}\frac{1}{\sin t}dt=\ln(\tan(\frac{t}{2}))|_{t_1}^{t_2}$. Similarly, for $\rho<0$, we have $t_1:=\cos^{-1}\left(\frac{-\rho}{\sqrt{\rho^2+4}}\right), t_2:=\cos^{-1}\left(\frac{\rho}{\sqrt{\rho^2+4}}\right)$, and the integral is the same one.
\end{enumerate}
}

{\bf Problem 2}. For each of the following paths $\gamma$ from $P=i$ to $Q=1+i$, draw the image of the path $\gamma$ in $\H^2$ and compute their hyperbolic lengths $\ell_{\H^2}(\gamma)$:

\begin{enumerate}
    \item $\gamma(t)=t+i$, $t\in[0,1]$.
    \item $\gamma(\theta)=\frac{1}{2} + \frac{\sqrt{5}}{2}(\cos\theta + i\sin\theta)$, where $\theta\in[\arccos\left(-\frac{1}{\sqrt{5}}\right),\arccos\left(\frac{1}{\sqrt{5}}\right)]$.
    \item $\gamma(t) = t + i \left( 4\left(t - \frac{1}{2}\right)^2 \right)$, $t \in [0, 1]$.
    \item Which of the above three paths has minimal hyperbolic length?\\
\end{enumerate}\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item We have $x(t)=t,y(t)=1, x'(t)=1,y'(t)=0$, from which we immediately compute $\ell_{\H^2}(\gamma)=\int_0^1\frac{\sqrt{1^2+0^2}}{1}dt=\int_0^1 1dt=1.$
    \item We have $x(\theta)=\frac12+\frac{\sqrt{5}}{2}\cos \theta, y(\theta)=\frac{\sqrt{5}}{2}\sin \theta$, and using the formula for computing the length of a geodesic segment, we have $\ell_{\H^2}(\gamma)=\ln\left(\tan\left(\frac{\theta}{2}\right)\right)\bigg|_{\cos^{-1}\left(-\frac{1}{\sqrt{5}}\right)}^{\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)}$.
    \item We have $x(t)=t,y(t)=4(t-\frac12)^2$, and $t\in[0,1]$. However, when $t=\frac12$, we get $y=0\not\in\H^2$. In particular, $\gamma$ reaches a limit point, and hence it would follow that $\gamma$ has length $\infty$.
    \item The second path has minimal hyperbolic length, since it is the geodesic segment from $P$ to $Q$. Here is the picture of them (\textcolor{purple}{purple} is (2),\textcolor{green}{green} is (1), \textcolor{red}{red} is (3)):
    \begin{figure}[htp]
        \centering
        \includegraphics[width=0.3\linewidth]{ProblemSets/paths gamma P Q.png}
    \end{figure}
\end{enumerate}
}

{\bf Problem 3} Consider the point $i\in\H^2$.

\begin{enumerate}
    \item Show that the set of points in $\H^2$ at hyperbolic unit distance from $i$ is
$$S_i:=\{z\in\H^2: |z-i\cosh(1)|=\sinh(1)\}.$$

\item Draw the set $S_i$ qualitatively in $\H^2$, explain why it is a circle, find its center and its radius.
\end{enumerate}
\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item We use the hyperbolic distance formula $d_{\H^2}(z,w)=\cosh^{-1}\left(1+\frac{|z-w|^2}{2\text{ Im}(z)\text{ Im}(w)}\right)$, and set $w=i, z=x+yi\in\H^2$. Then, we have
    \begin{align*}
        d_{\H^2}(z,i)=1&\iff 1=\cosh^{-1}\left(1+\frac{|z-i|^2}{2\text{ Im}(z)}\right)=\cosh^{-1}\left(1+\frac{|x+yi-i|^2}{2y}\right)\\
        &\iff \cosh(1)=1+\frac{x^2+(y-1)^2}{2y}\\
        &\iff 2y(\cosh(1)-1)=x^2+(y-1)^2=x^2+y^2-2y+1\\
        &\iff x^2+y^2-2y\cosh(1)+1=0\\
        &\iff x^2+(y-\cosh(1))^2-(\cosh(1))^2+1=0\\
        &\iff x^2+(y-\cosh(1))^2=(\cosh(1))^2-1
    \end{align*}
    Using the identity $\cosh^2(x)-\sinh^2(x)=1\iff \sinh^2(x)=\cosh^2(x)-1$, and noting that $x^2+(y-\cosh(1))^2=|x+i(y-\cosh(1))|^2=|z-i\cosh(1)|^2$, we get
    $|z-i\cosh(1)|^2=\sinh^2(1),$ and thus $|z-i\cosh(1)|=\sinh(1)$.
    \item Since $\cosh(1),\sinh(1)$ are constants, $|z-i\cosh(1)|=\sinh(1)$ precisely maps out a circle centered at $i\cosh(1)\approx 1.543i$ with radius $\sinh(1)\approx 1.175$. \begin{figure}[htp]
        \centering
        \includegraphics[width=0.3\linewidth]{ProblemSets/unit circle about i.png}
    \end{figure}
\end{enumerate}
}

\section{Problems on hyperbolic lines}

{\bf Problem 4}. Consider $1+i\in\H^2$ and $L:=\{z\in\H^2:\mbox{Re}(z)=1\}$ a hyperbolic line containing it. Given the point $i\in\H^2$ outside $L$, show that there are infinitely many hyperbolic lines through $i$ that are parallel to $L$.\\

\textcolor{blue}{
This is essentially the same problem as PSet 5 problem 2. See solution there.
}\\

{\bf Problem 5}. Consider two points $P=iy_1$ and $Q=iy_2$ in $\H^2$, $y_1,y_2\in\R_{>0}$ and $y_1<y_2$.

\begin{enumerate}
\item Show that the unique hyperbolic line $E(P,Q)\sse\H^2$ given as the set of equidistant points to $P$ and $Q$ is
$$E(P,Q)=\{z\in\H^2: |z|^2=y_1y_2\}.$$

\item Describe explicitly, using a formula, a hyperbolic isometry $f:\H^2\lr\H^2$ whose fixed point set coincides with $E(P,Q)$.\\
\end{enumerate}
\textcolor{blue}{
\begin{enumerate}
    \item By a Theorem from class, we know that the set of equidistant points to two points is a hyperbolic line. For the case $P=iy_1,Q=iy_2$, it is clear by symmetry that the geodesic must be centered at $0$. Then, we find the point on the imaginary axis equidistant to $P,Q$, and use this point to determine the radius. Suppose $z_0=iy\in\H^2$ such that $d_{\H^2}(iy,iy_1)=d_{\H^2}(iy,iy_2),$ with $y_1<y<y_2$. By a Theorem from class, we also know that the hyperbolic distance between two points $(x,y_1),(x,y_2)$ with the same real component is $\ln\left(\frac{y_2}{y_1}\right)$. Thus, we have $\ln\left(\frac{y}{y_1}\right)=\ln\left(\frac{y_2}{y}\right)\iff \frac{y}{y_1}=\frac{y_2}{y}\iff y^2=y_1y_2\iff y=\sqrt{y_1y_2}$. In particular, the Euclidean distance from $0$ to $z_0$ is $y=\sqrt{y_1y_2}$, and this is the radius of the geodesic. Therefore, the set of equidistant points to $P,Q$ is the geodesic centered at $0$ with radius $\sqrt{y_1y_2}$, and this is precisely $E(P,Q)$. 
    %We use the hyperbolic distance formula $d_{\H^2}(z,w)=\cosh^{-1}\left(1+\frac{|z-w|^2}{2\text{ Im}(z)\text{ Im}(w)}\right)$. Assuming $z=x+yi\in\H^2$ is equidistant from $P,Q$, we have:
    %\begin{align*}
    %    d_{\H^2}(z,P)=d_{\H^2}(z,Q)&\iff 1+\frac{|z-iy_1|^2}{2yy_1}=1+\frac{|z-iy_2|^2}{2yy_2}\\
    %    &\iff \frac{x^2+(y-y_1)^2}{y_1}=\frac{x^2+(y-y_2)^2}{y_2}\\
    %    &\iff x^2y_2+(y^2-2yy_1+y_1^2)y_2=x^2y_1+(y^2-2yy_2+y_2^2)y_1\\
    %    &\iff x^2y_2-x^2y_1+y^2y_2=y^2y_2=y_1y_2^2-y_1^2y_2\\
    %    &\iff (x^2+y^2)(y_2-y_1)=y_1y_2(y_2-y_1)\\
    %    &\iff y_1y_2=x^2+y^2=|z|^2,
    %\end{align*}
    %so the set of equidistant points to $P$ and $Q$ is $E(P,Q)$.
    \item We want to find the formula for reflection across $E(P,Q)$. Recall the standard inversion across the unit circle $|z|=1$ is $I(z)=\frac{1}{\overline{z}}$, so we do a conjugation by $d_{\sqrt{y_1y_2}}(z)=\sqrt{y_1y_2}\cdot z, d_{\sqrt{y_1y_2}}^{-1}(z)=\frac{z}{\sqrt{y_1y_2}}$ ($d_{\sqrt{y_1y_2}}^{-1}$ scales $E(P,Q)$ to align with the unit circle, and $d_{\sqrt{y_1y_2}}$ scales the unit circle to align with $E(P,Q)$), and therefore the reflection across $E(P,Q)$ is 
    $$r_E(z)=d_{\sqrt{y_1y_2}}\circ I\circ d_{\sqrt{y_1y_2}}^{-1}(z)=d_{\sqrt{y_1y_2}}\circ I\left(\frac{z}{\sqrt{y_1y_2}}\right)=d_{\sqrt{y_1y_2}}\left(\frac{\sqrt{y_1y_2}}{\overline{z}}\right)=\frac{y_1y_2}{\overline{z}}.$$
\end{enumerate}
}

{\bf Problem 6}. Consider the two hyperbolic lines $L_1:=\{z\in\H^2:|z-2|=2\}$ and $L_2:=\{z\in\H^2:|z+2|=2\}$. Let $\rho\in(0,1]$ be a given real number.

\begin{enumerate}
\item Find the coordinates of the point $z_1(\rho)\in L_1$, given as the unique point in $L_1$ with $x$-coordinate equal to $\rho$. Similarly, find the coordinates of the point $z_2(\rho)\in L_2$ given as the unique point in $L_2$ with $x$-coordinate $-\rho$.\\

\item Show that the hyperbolic distance between $z_1(\rho)$ and $z_2(\rho)$ is $\ln\left(\frac{2+\sqrt{\rho}}{2-\sqrt{\rho}}\right)$.\\

\item Conclude that two hyperbolic lines sharing a common vertex in $\R\sse \partial\overline{\H}^2$ get closer and closer to each other in Euclidean distance with a rate of $\sqrt{\rho}$ when $\rho\to0$.
\end{enumerate}
\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item $L_1=\{(x,y)\in\H^2:(x-2)^2+y^2=4\}$, so $z_1(\rho)=\rho+yi$ such that $(\rho-2)^2+y^2=4\implies y=\sqrt{4-(\rho-2)^2}$. Similarly for $z_2(\rho)$, we get $z_2(\rho)=-\rho+yi$ such that $(2-\rho)^2+y^2=4\implies y=\sqrt{4-(\rho-2)^2}$. Thus, $z_1(\rho)=\rho+i\sqrt{4-(\rho-2)^2}, z_2(\rho)=-\rho+i\sqrt{4-(\rho-2)^2}$.
    \item First, we find the general formula for distance between $(\alpha,\beta),(-\alpha,\beta)$ for $\alpha,\beta>0$. The circle geodesic through them is centered at $(0,0)$ with radius $\sqrt{\alpha^2+\beta^2}$, so the equation is $x^2+y^2=\alpha^2+\beta^2$. Then, we have $y=\sqrt{\alpha^2+\beta^2-x^2}, \frac{dy}{dx}=\frac{-x}{\sqrt{\alpha^2+\beta^2-x^2}}$, and thus $$
    d_{\H^2}((\alpha,\beta),(-\alpha,\beta))=\int_{-\alpha}^{\alpha}\frac{\sqrt{1+\frac{x^2}{\alpha^2+\beta^2-x^2}}}{\sqrt{\alpha^2+\beta^2-x^2}}dx=\int_{-\alpha}^\alpha \frac{\sqrt{\alpha^2+\beta^2}}{\alpha^2+\beta^2-x^2}dx.$$
    Therefore, $d_{\H^2}(z_1(\rho),z_2(\rho))=\int_{-\rho}^\rho \frac{\sqrt{\rho^2+4-(\rho-2)^2}}{\rho^2+4-(\rho-2)^2-x^2}dx=\int_{-\rho}^\rho \frac{\sqrt{4\rho}}{4\rho-x^2}dx=\int_{-\rho}^\rho \frac{2\sqrt{\rho}}{4\rho-x^2}dx$. Then, we apply partial fractions:
    \begin{align*}
        \int_{-\rho}^\rho \frac{2\sqrt{\rho}}{4\rho-x^2}dx&=\int_{-\rho}^\rho \frac{A}{2\sqrt{\rho}-x}+\frac{B}{2\sqrt{\rho}+x}dx\\
        &\implies 2\sqrt{\rho}=2\sqrt{\rho}A+Ax+2\sqrt{\rho}B-Bx\implies A=B=\frac12\\
        \int_{-\rho}^\rho \frac{2\sqrt{\rho}}{4\rho-x^2}dx&=\int_{-\rho}^\rho \frac{1/2}{2\sqrt{\rho}-x}+\frac{1/2}{2\sqrt{\rho}+x}dx\\
        &=\left(\frac12 (-\ln(2\sqrt{\rho}-x)+\ln(2\sqrt{\rho}+x)\right)\bigg|_{-\rho}^\rho\\
        &=\left(\frac12\ln\left(\frac{2\sqrt{\rho}+x}{2\sqrt{\rho}-x}\right)\right)\bigg|_{-\rho}^\rho\\
        &=\frac12\left(\ln\left(\frac{2\sqrt{\rho}+\rho}{2\sqrt{\rho-\rho}}\right)-\ln\left(\frac{2\sqrt{\rho}-\rho}{2\sqrt{\rho}+\rho}\right)\right)\\
        &=\frac12\left(\ln\left(\frac{2\sqrt{\rho}+\rho}{2\sqrt{\rho-\rho}}\right)+\ln\left(\frac{2\sqrt{\rho}+\rho}{2\sqrt{\rho-\rho}}\right)\right)\\
        &=\ln\left(\frac{2\sqrt{\rho}+\rho}{2\sqrt{\rho-\rho}}\right)=\ln\left(\frac{2+\sqrt{\rho}}{2-\sqrt{\rho}}\right),
    \end{align*}
    as desired.
    \item We know that the limit
    $$\lim_{\rho\to0}\ln\left(\frac{2+\sqrt{\rho}}{2-\sqrt{\rho}}\right)=0$$
    vanishes. The question is asking how fast it vanishes, i.e.~the rate at which it converges to $0$. To compute that rate, and show it is equal to $\sqrt{\rho}$, we compute the Taylor (a.k.a.~Maclaurin) series expansion expansion of $\ln\left(\frac{2+\sqrt{\rho}}{2-\sqrt{\rho}}\right)$ around $\rho = 0$. Since the Taylor series expansion for $\ln\left(\frac{1+x}{1-x}\right)$ for $|x| < 1$ is
\begin{equation*}
    \ln\left(\frac{1+x}{1-x}\right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \dots \right),
\end{equation*}
we re-write
$$\frac{2+\sqrt{\rho}}{2-\sqrt{\rho}} = \frac{2\left(1 + \frac{\sqrt{\rho}}{2}\right)}{2\left(1 - \frac{\sqrt{\rho}}{2}\right)} = \frac{1 + \frac{\sqrt{\rho}}{2}}{1 - \frac{\sqrt{\rho}}{2}}.$$
Therefore we obtain
\begin{equation*}
    \ln\left(\frac{1 + \frac{\sqrt{\rho}}{2}}{1 - \frac{\sqrt{\rho}}{2}}\right) = 2 \left( \left(\frac{\sqrt{\rho}}{2}\right) + \frac{1}{3}\left(\frac{\sqrt{\rho}}{2}\right)^3 + \frac{1}{5}\left(\frac{\sqrt{\rho}}{2}\right)^5 + \frac{1}{7}\left(\frac{\sqrt{\rho}}{2}\right)^7 + \dots \right)
\end{equation*}
This expression can be simplified to
\begin{align*}
    \ln\left(\frac{1 + \frac{\sqrt{\rho}}{2}}{1 - \frac{\sqrt{\rho}}{2}}\right) &= 2 \left( \frac{\rho^{1/2}}{2} + \frac{1}{3} \cdot \frac{\rho^{3/2}}{8} + \frac{1}{5} \cdot \frac{\rho^{5/2}}{32} + \frac{1}{7} \cdot \frac{\rho^{7/2}}{128} + \dots \right) \\
    &= \rho^{1/2} + \frac{2}{24}\rho^{3/2} + \frac{2}{160}\rho^{5/2} + \frac{2}{896}\rho^{7/2} + \dots\\
    &=
   \sqrt{\rho} + \frac{1}{12}\rho^{3/2} + \frac{1}{80}\rho^{5/2} + \frac{1}{448}\rho^{7/2} + \mathcal{O}(\rho^{9/2}).
\end{align*}
Thus the convergence rate near $\rho\to0$ is indeed dominated by $\sqrt{\rho}$, as required.
\end{enumerate}
}

{\bf Problem 7}. Consider the three hyperbolic lines $L_1:=\{z\in\H^2:|z-1|=1\}$, $L_2:=\{z\in\H^2:|z+1|=1\}$ and $L_3:=\{z\in\H^2:|z|=2\}$. Show that the area of the hyperbolic triangle bounded by $L_1,L_2,L_3$ is $\pi$.

\textcolor{blue}{
A Theorem from class states that the area of a hyperbolic triangle with angles $\alpha,\beta,\gamma$ is $\pi-\alpha-\beta-\gamma$. So to show this triangle has area $\pi$, it suffices to show that the angles $\alpha,\beta,\gamma$ are all equal to $0$, and this happens precisely when the lines meet on $\partial \H^2$ (this is because the angle a geodesic makes with $\partial \H^2$ is always $\pi/2$ by construction, so when two different geodesics meet at $\partial \H^2$, they both make an angle $\pi/2$ with $\partial \H^2$, and hence the angle between the two geodesics must be $0$). Note that $L_1$ has endpoints $0,2$, $L_2$ has endpoints $-2,0$, and $L_3$ has endpoints $2,-2$, and therefore the angles are all equal to $0$, hence the area of this triangle is $\pi$. Here's what the triangle looks like:
\begin{figure}[htp]
    \centering
    \includegraphics[width=0.3\linewidth]{ProblemSets/hyp triangle.png}
\end{figure}
}

\section{Problems on hyperbolic isometries}

{\bf Problem 8}. Consider the two maps $f_1,f_2:\H^2\lr\H^2$ given by
$$f_1(z)=-\overline{z},\quad f_2(z)=\frac{1}{\overline{z}}$$

\begin{enumerate}
    \item Show that $f_1$ and $f_2$ are orientation-reversing hyperbolic isometries.\\

    \item Describe the fixed points of $f_1$ and the fixed points of $f_2$.\\

    \item Determine whether $f_1\circ f_2$ equals $f_2\circ f_1$ or not.\\

    \item Find all the fixed points of $f_1\circ f_2$.\\

    \item Show that $f_1\circ f_2$ is not the identity but $(f_1\circ f_2)^2$ is the identity.\\
\end{enumerate}\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item It suffices to write $f_1,f_2$ in the form $f(z)=\frac{-a\overline{z}+b}{-c\overline{z}+d}$, with $ad-bc=1$, since we know from a theorem that the orientation reversing isometries on $\H^2$ are of this form. So, we have $f_1(z)=\frac{-\overline{z}+0}{-0\overline{z}+1}$, $f_2(z)=\frac{-0\overline{z}+1}{-(-1)\overline{z}+0}$, and these already satisfy $ad-bc=1$, hence they are orientation-reversing hyperbolic isometries.
    \item We've seen in discussion (or maybe also in class) that $f_1(z)=-\overline{z}$ is the reflection across the $y$-axis, hence the fixed points of $f_1$ are all the points on the $y$-axis. We've also seen in discussion that $f_2(z)$ is the inversion across the unit circle, hence the fixed points of $f_2$ are all the points on the unit circle $|z|=1$.
    \item Note that the $y$-axis and the unit circle intersect at an angle of $\pi/2$, thus $f_1\circ f_2$ and $f_2\circ f_1$ are both rotations by $\pi$. We can also do this with explicit computations:
    \begin{align*}
        f_1\circ f_2(z)&=f_1\left(\frac{1}{\overline{z}}\right)=f_1\left(\frac{z}{|z|^2}\right)=-\frac{\overline{z}}{|z|^2}=-\frac{\overline{z}}{\overline{z}z}=-\frac{1}{z}\\
        f_2\circ f_1(z)&=f_2(-\overline{z})=\frac{1}{\overline{-\overline{z}}}=-\frac{1}{z}.
    \end{align*}
    \item $f_1\circ f_2$ is the rotation about $i$ by $\pi$, hence there is a single fixed point of $f_1\circ f_2$, namely $i$.
    \item Since $f_1\circ f_2(\frac{i}{2})=-\frac{1}{i/2}=-\frac{2}{i}=2i\neq \frac{i}{2}$, it follows that $f_1\circ f_2\neq $ Id. However, we have $(f_1\circ f_2)^2(z)=(f_1\circ f_2)(-\frac{1}{z})=-\frac{1}{-1/z}=z$, hence $(f_1\circ f_2)^2=$ Id.
\end{enumerate}
}

{\bf Problem 9}. Consider the map
$$f:\H^2\lr\H^2,\quad f(z)=\frac{3\overline{z}+5}{5\overline{z}+3}.$$

\begin{enumerate}
    \item Show that $f$ is an orientation-reversing hyperbolic isometry.\\

    \item Prove that $f$ has no fixed points.\\

    \item Find a hyperbolic line $L\sse\H^2$ invariant under $f$, i.e.~such that $f(L)=L$.\\

    \item Find an explicit formula for the unique hyperbolic isometry $g:\H^2\lr\H^2$ such that $f\circ g$ is the identity.\\

    \item Find the fixed points of $g$.
\end{enumerate}

\textcolor{blue}{
\begin{enumerate}
    \item As with problem 8, we want to write $f$ in the form $f(z)=\frac{-a\overline{z}+b}{-c\overline{z}+d}$, with $ad-bc=1$, and then it would follow from a theorem in class that $f$ is orientation-reversing. For $f(z)=\frac{3\overline{z}+5}{5\overline{z}+3}$, we have $ad-bc=-9-(-25)=16$, and we want to normalize by dividing by $\sqrt{16}=4$. So $f(z)=\frac{\frac34\overline{z}+\frac54}{\frac54\overline{z}+\frac34}$, and here $ad-bc=1$, hence it is an orientation-reversing hyperbolic isometry.
    \item Assume $z=x+yi\in\H^2$ is fixed by $f$. Then
    \begin{align*}
        &x+yi=z=\frac{3\overline{z}+5}{5\overline{z}+3}=\frac{3x+5-3yi}{5x+3-5yi}\\
        &\iff (x+yi)(5x+3-5yi)=3x+5-3yi\\
        &\iff 5x^2+3x-5xyi+5xyi+3yi+5y^2=3x+5-3yi\\
        &\iff 5x^2+5y^2=5 \text{ and } 3yi=-3yi,
    \end{align*}
    But the latter implies $y=0$, hence $z\notin \H^2$. So $f$ has no fixed points.
    \item To find a hyperbolic line invariant under $f$, we look for the two endpoints of the line on $\partial \H^2$. If $x\in\R\subseteq\partial \H^2$ is fixed by $f$, then we have $x=\frac{3x+5}{5x+3}\implies 5x^2+3x=3x+5\implies x^2=1\implies x=\pm 1$. So the geodesic through $1,-1$ is invariant under $f$, i.e. the unit circle $|z|=1$.
    \item We find the inverse of $f(z)=\frac{\frac34\overline{z}+\frac54}{\frac54\overline{z}+\frac34}.$ Define $\widehat{f}(z):=\frac{-\frac34z+\frac54}{-\frac54z+\frac34}$, and consider the reflection across the $y$-axis $r(z)=-\overline{z}$. We have that $f=\widehat{f}\circ r$, and thus $g=f^{-1}=(\widehat{f}\circ r)^{-1}=r\circ \widehat{f}^{-1}$. Using matrix representations, $\widehat{f}\longleftrightarrow \begin{bmatrix}
        -\frac34 & \frac54\\
        -\frac54 & \frac34
    \end{bmatrix}$, and its inverse matrix is $\begin{bmatrix}
        \frac34 & -\frac54\\
        \frac54 & -\frac34
    \end{bmatrix}\longleftrightarrow \widehat{f}^{-1}(z):=\frac{\frac34z-\frac54}{\frac54z-\frac34}$. Then, we compute 
    $$g(z)=r\circ \widehat{f}^{-1}(z)=-\overline{\left(\frac{\frac34z-\frac54}{\frac54z-\frac34}\right)}=-\frac{\overline{\frac34z-\frac54}}{\overline{\frac54z-\frac34}}=-\frac{\frac34\overline{z}-\frac54}{\frac54\overline{z}-\frac34}=\frac{-\frac34\overline{z}+\frac54}{\frac54\overline{z}-\frac34}.$$
    \item If $z_0$ is fixed by $g$, then $f\circ g(z_0)=f(z_0)\neq z_0$ because we showed in part (2) that $f$ has no fixed points. But by part (4), $f\circ g=$ Id, thus we must have $f\circ g(z_0)=z_0$, which is a contradiction. Therefore, $g$ has no fixed points as well. Alternatively, we can also compute explicitly for fixed points in the same way as part (2).
\end{enumerate}
}

{\bf Problem 10}. Consider the map
$$f:\H^2\lr\H^2,\quad f(z)=\frac{(\sqrt{3}-1)z + 2}{-z + (\sqrt{3}+1)}.$$

\begin{enumerate}
    \item Show that $1+i$ is the unique fixed point of $f$.\\

    \item Show that $f^6$ is the identity.\\

    \item Prove that $f$ is a hyperbolic rotation centered at $1+i$ and determine its angle.
\end{enumerate}\vspace{-10mm}
\textcolor{blue}{
\begin{enumerate}
    \item Suppose $f(z_0)=z_0$. Then we have
    \begin{align*}
        &z=\frac{(\sqrt{3}-1)z+2}{-z+(\sqrt{3}+1)}\\
        &\iff -z^2+\sqrt{3}z+z=\sqrt{3}z-z+2\\
        &\iff z^2-2z+2=0\\
        &\iff z=\frac{2\pm\sqrt{4-8}}{2}=\frac{2\pm 2i}{2}=1\pm i,
    \end{align*}
    and since $1-i\notin\H^2$, it follows that the unique fixed point of $f$ is $1+i$.
    \item We have $f\longleftrightarrow \begin{bmatrix}
        \sqrt{3}-1&2\\
        -1&\sqrt{3}+1
    \end{bmatrix},$ and we just compute $f^6$ directly:
    \begin{align*}
        \begin{bmatrix}
        \sqrt{3}-1&2\\
        -1&\sqrt{3}+1
    \end{bmatrix}^6&=\left(\begin{bmatrix}
        \sqrt{3}-1&2\\
        -1&\sqrt{3}+1
    \end{bmatrix}^3\right)^2\\
    &=\left(\begin{bmatrix}
        \sqrt{3}-1&2\\
        -1&\sqrt{3}+1
    \end{bmatrix}\begin{bmatrix}
        3-2\sqrt{3}+1-2 & 2\sqrt{3}-2+2\sqrt{3}+2\\
        -\sqrt{3}+1-\sqrt{3}-1 & -2+3+2\sqrt{3}+1
    \end{bmatrix}\right)^2\\
    &=\left(\begin{bmatrix}
        \sqrt{3}-1&2\\
        -1&\sqrt{3}+1
    \end{bmatrix}\begin{bmatrix}
        2-2\sqrt{3} & 4\sqrt{3}\\
        -2\sqrt{3} & 2+2\sqrt{3}
    \end{bmatrix}\right)^2\\
    &=\left(\begin{bmatrix}
        2\sqrt{3}-6-2+2\sqrt{3}-4\sqrt{3} & 12-4\sqrt{3}+4+4\sqrt{3}\\
        -2+2\sqrt{3}-6-2\sqrt{3} & -4\sqrt{3}+2\sqrt{3}+6+2+2\sqrt{3}
    \end{bmatrix}\right)^2\\
    &=\left(\begin{bmatrix}
        -8 & 16\\
        -8 & 8
    \end{bmatrix}\right)^2=\left(8\begin{bmatrix}
        -1 & 2\\
        -1 & 1
    \end{bmatrix}\right)^2\\
    &=64\begin{bmatrix}
        -1 & 0\\
        0 & -1
    \end{bmatrix},
    \end{align*}
    which we can just normalize to get $\begin{bmatrix}
        1 & 0\\
        0 & 1
    \end{bmatrix}$. Hence, $f^6=$ Id.
    \item Since $f$ is an orientation-preserving isometry, and fixes a single point in $\H^2$, it follows from the classification of M\"obius Transformations that $f$ is a rotation about its fixed point $1+i$. To determine the angle of rotation, we consider the geodesic $L:=\{(x,y)\in\H^2:x=1\}$, and observe how it behaves under $f$. Note that $f(1)=\frac{\sqrt{3}+1}{-1+\sqrt{3}+1}=\frac{3+\sqrt{3}}{3}=1+\frac{\sqrt{3}}{3}$, and $f(\infty)=\frac{\sqrt{3}-1}{-1}=1-\sqrt{3}$. So $f(L)$ is the geodesic with endpoints at $1+\frac{\sqrt{3}}{3},1-\sqrt{3}$, center $\frac{1+\frac{\sqrt{3}}{3}+1-\sqrt{3}}{2}=\frac{2-\frac{2\sqrt{3}}{3}}{2}=1-\frac{\sqrt{3}}{3}$ and radius $\frac{1+\frac{\sqrt{3}}{3}-1+\sqrt{3}}{2}=\frac{2\sqrt{3}}{3}$, so its equation is
    $$\left\{(x,y)\in\H^2:\left(x-1+\frac{\sqrt{3}}{3}\right)^2+y^2=\frac43\right\}.$$ We want to find the equation of the tangent line at $1+i$, and find the angle between this line and the line $x=1$. We consider the vector $\frac{\sqrt{3}}{3},1\rangle$ (this is the vector from the center of the circle to $1+i$), and note that the tangent line is perpendicular to this vector, hence for any point $(x,y)$ on the tangent line, we must have $\langle\frac{\sqrt{3}}{3},1\rangle\cdot \langle x-1,y-1\rangle=0\iff \frac{\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}+y-1=0\iff y=-\frac{\sqrt{3}}{3} x+1+\frac{\sqrt{3}}{3}$. Consider the Euclidean right triangle bounded by $L$, the tangent line, and the $x$-axis. We note that the tangent line intersects the $x$-axis at $(\sqrt{3}+1,0)$, and that the other two vertices are $(1,0),(1,1)$. This forms a special triangle ($30^\circ -60^\circ-90^\circ$) triangle, and thus the angle formed by $L$ and the tangent line is $\frac{\pi}{3}$. It follows that its vertical angle is also $\frac{\pi}{3}$, and hence $f$ is a rotation about $1+i$ by angle $\frac{\pi}{3}$.
    \begin{figure}[htp]
        \centering
        \includegraphics[width=0.3\linewidth]{ProblemSets/rot about 1,1.png}
    \end{figure}\\
    Alternatively, we could also do a conjugation by the limit rotation $t_1(z)=z+1$ and look at $t_1^{-1}\circ f\circ t_1$, which is a rotation about $i$. We compute this formula directly:
    \begin{align*}
        t_1^{-1}\circ f\circ t_1(z)&=t_1^{-1}\circ f(z+1)\\
        &=t_1^{-1}\left(\frac{(\sqrt{3}-1)z+\sqrt{3}-1+2}{-z-1+\sqrt{3}+1}\right)\\
        &=\frac{(\sqrt{3}-1)z+\sqrt{3}+1}{-z+\sqrt{3}}-\frac{-z+\sqrt{3}}{-z+\sqrt{3}}\\
        &=\frac{\sqrt{3}z+1}{-z+\sqrt{3}}\longleftrightarrow\begin{bmatrix}
            \sqrt{3}&1\\
            -1&\sqrt{3}
        \end{bmatrix},
    \end{align*}
    which has determinant $4$, and we can normalize to get $t_1^{-1}\circ f\circ t_1\longleftrightarrow\begin{bmatrix}
        \frac{\sqrt{3}}{2}&\frac12\\
        -\frac12&\frac{\sqrt{3}}{2}
    \end{bmatrix}.$ Recall we've seen in the homework that the rotation about $i$ by angle $\theta$ is given by the matrix $\begin{bmatrix}
        \cos \theta/2 &\sin\theta/2\\
        -\sin\theta/2&\cos\theta/2
    \end{bmatrix}.$ By observation, we have that $\cos \pi/6=\frac{\sqrt{3}}{2},\sin \pi/6=\frac{1}{2}$, and hence $t_1^{-1}\circ f\circ t_1\longleftrightarrow\begin{bmatrix}
        \frac{\sqrt{3}}{2}&\frac12\\
        -\frac12&\frac{\sqrt{3}}{2}
    \end{bmatrix}=\begin{bmatrix}
        \cos \pi/6 &\sin\pi/6\\
        -\sin\pi/6&\cos\pi/6
    \end{bmatrix}$, and therefore $t_1^{-1}\circ f\circ t_1$ is the rotation about $i$ by angle $\pi/3$. We conclude that $f$ is the rotation about $1+i$ by angle $\pi/3$.
\end{enumerate}
}

\end{document}