\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Consider the two circles
\[ A_1 = \{ z\in {\bf C}\: |\: |z - (2 + 2i)| = 4\} \]
and
\[ A_2 = \{ z\in {\bf C}\: |\: |z - 2| = 3\}. \]
Find the points of intersection of these two circles, and
determine the angles between the two circles by determining the
angle between their tangent lines at one of the points of
intersection, call it $\xi$. (Which point of intersection you use
is your choice.)

Consider now the M\"obius transformation $m(z) =
\ds\frac{z}{z+1}$. Determine the equations of the image circles
$m(A_1)$ and $m(A_2)$. Determine the angle between $m(A_1)$ and
$m(A_2)$ at $m(\xi)$ by determining the angle between their
tangent lines.

Confirm that the angle between $A_1$ and $A_2$ at $\xi$ is equal
to the angle between $m(A_1)$ and $m(A_2)$ at $m(\xi)$.

\medskip

{\bf Answer}

Start by determining the intersection points of $A_1$ and $A_2$:

$A_1$ is given by $|z-(2+2i)|=4$ which expands out to

$$z\bar z-(2+2i)\bar z-(2-2i)z-8=0$$

\bigskip

$A_2$ is given by $|z-2|=3$ which expands out to

$$z\bar z-2\bar z-2z-5=0$$

\bigskip

$z \bar z-(2+2i)\bar z-(2-2i)z-8=z\bar z-2\bar z-2z-5$

simplifies to $-2i\bar z+2iz-3=0$ which gives (setting $z=x+iy$)
$y=\ds\frac{-3}{4}$.

\bigskip

Plugging back into either (both) of the equations for the circles
$A_1A_2$ we get $x=2 \pm \ds\frac{\sqrt{135}}{4}$.

\bigskip

So $A_1A_2$ intersect at $\un{\left(2 \pm
\ds\frac{\sqrt{135}}{4}\right)-\ds\frac{3}{4}i}$

\bigskip

Work at $\un{z_0=\left(2 \pm
\ds\frac{\sqrt{135}}{4}\right)-\ds\frac{3}{4}i}$

\bigskip

\un{The radius of $A_1$} passes through $z_0$ and $2+2i$ and so
the slope of the radius is
$m_1=\ds\frac{2-(-\frac{3}{4})}{2-(2+\frac{\sqrt{135}}{4})}=
\ds\frac{\frac{11}{4}}{-\frac{\sqrt{135}}{4}}=\ds\frac{-11}{\sqrt{135}}$

\bigskip

The \un{slope of the tangent line is $M_1=\ds\frac{-1}{m_1}$} and
so the equation of the tangent line is:

$y+\ds\frac{3}{4}=\ds\frac{\sqrt{135}}{11}\left(x-\left(2+\ds\frac{\sqrt{135}}{4}\right)\right)$

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$\ds\frac{1}{2i}(z-\bar
z)+\ds\frac{3}{4}=\ds\frac{\sqrt{135}}{11}\left(\ds\frac{1}{2}(z+\bar
z)-\left(2+\ds\frac{\sqrt{135}}{4}\right)\right)$

\bigskip

$\ds\frac{-i}{2}z+\ds\frac{i}{2}\bar z
+\ds\frac{3}{4}=\ds\frac{\sqrt{135}}{22}z+\ds\frac{\sqrt{135}}{22}\bar
z-\ds\frac{\sqrt{135}}{11}\left(2+\ds\frac{\sqrt{135}}{4}\right)$

\bigskip

$0=\left(\ds\frac{\sqrt{135}}{22}+\ds\frac{i}{2}\right)z+
\left(\ds\frac{\sqrt{135}}{22}-\ds\frac{i}{2}\right)\bar
z-\ds\frac{2\sqrt{135}}{11}-\ds\frac{135}{44}-\ds\frac{3}{4}$

\bigskip

$\un{0=\left(\ds\frac{\sqrt{135}}{22}+\ds\frac{i}{2}\right)z+
\left(\ds\frac{\sqrt{135}}{22}-\ds\frac{i}{2}\right)\bar
z-\ds\frac{(8\sqrt{135}+168)}{44}}$

\bigskip

$0=\beta_1z+\bar \beta_1 \bar z +\gamma_1=0$

\bigskip

\un{Facts to note}: the angle that the tangent line to $A_1$ at
$z_0$ makes with $\bf{R}$ is $\theta_1=\arctan(M_1) \Rightarrow
\theta_1=0.8128$.

\bigskip

$\beta_1=\ds\frac{\sqrt{135}}{22}+\ds\frac{1}{2}$

\bigskip

$\gamma_1=-\ds\frac{(8\sqrt{135}+168)}{44}$

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\un{The radius of $A_2$} passes through $z_0$ and $2$ and so the
slope of the radius is
$m_2=\ds\frac{0-(-\frac{3}{4})}{2-(2+\frac{\sqrt{135}}{4})}=
\ds\frac{\frac{3}{4}}{-\frac{\sqrt{135}}{4}}=\ds\frac{-1}{\sqrt{15}}$


\newpage
The \un{slope of the tangent line is
$M_2=\ds\frac{-1}{m_2}=\sqrt{15}$} and so the equation of the
tangent line is:

$y+\ds\frac{3}{4}=\sqrt{15}\left(x-\left(2+\ds\frac{\sqrt{135}}{4}\right)\right)$

\bigskip

$y+\ds\frac{3}{4}=\sqrt{15}x-2\sqrt{15}-\ds\frac{\sqrt{15} \cdot
\sqrt{135}}{4}$

\bigskip

$\ds\frac{-i}{2}(z-\bar
z)+\ds\frac{3}{4}=\sqrt{15}\left(\ds\frac{1}{2}(z+\bar
z)\right)-2\sqrt{15}-\ds\frac{45}{4}$

\bigskip

$\left(\ds\frac{\sqrt{15}}{2}+\ds\frac{i}{2}\right)z+
\left(\ds\frac{\sqrt{15}}{2}-\ds\frac{i}{2}\right)\bar
z-2\sqrt{15}-\ds\frac{45}{4}-\ds\frac{3}{4}=0$

\bigskip

$\left(\ds\frac{\sqrt{15}}{2}+\ds\frac{i}{2}\right)z+
\left(\ds\frac{\sqrt{15}}{2}-\ds\frac{i}{2}\right)\bar
z-2\sqrt{15}-12=0$

\bigskip

$0=\beta_2z+\bar \beta_2 \bar z +\gamma_2=0$

\bigskip

\un{$\beta_2=\ds\frac{\sqrt{15}}{2}+\ds\frac{i}{2},\ \
\gamma_2=-2\sqrt{15}-12$}

\bigskip

The angle that the tangent line to $A_2$ at $z_0$ makes with
$\bf{R}$ is

$\theta_2=\arctan(M_2) \Rightarrow \theta_2=1.3181$.

\bigskip

So, the angle between $A_2$ and $A_1$ at $z_0$ is
\un{$\theta_2-\theta_1=0.5053$}.

\bigskip

Equation of $m$ (tangent line to $A_k$):

$w=m(z)=\ds\frac{z}{z+1},\ \ z=\ds\frac{w}{1-w}$

\bigskip

\begin{eqnarray*} 0 & = &
\beta_k z+\bar{\beta_k}\bar{z}+\gamma_k=0\\ & = & \beta_k
\ds\frac{w}{1-w}+\bar{\beta_k}\ds\frac{\bar{w}}
{1-\bar{w}}+\gamma_k\\ 0 & = & \beta_k
w(1-\bar{w})+\bar{\beta_k}\bar{w}(1-w)+ \gamma_k(1-w)(1-\bar{w})\\
0 & = & \beta_k
w-\beta_kw\bar{w}+\bar{\beta_k}\bar{w}-\bar{\beta_k}
w\bar{w}+\gamma_k-\gamma_kw -\gamma_k\bar{w}+\gamma_kw\bar{w}\\ 0
& = & (\gamma_k-\beta_k-\bar{\beta_k})w\bar{w}+(\beta_k-\gamma_k)w
+(\bar{\beta_k}-\gamma_k)\bar{w}+\gamma_k \end{eqnarray*}

\newpage
\un{Center of $m$(tangent line to $A_k$ at $z_0$)}:

$$\ds\frac{\gamma_k-\bar{\beta_k}}{\gamma_k-\beta_k-\bar{\beta_k}}=
\ds\frac{\gamma_k-\bar{\beta_k}}{\gamma_k-2\rm{Re}\beta_k}$$

\bigskip

$\un{k=1}$ $\gamma_1=-\ds\frac{(8\sqrt{135}+168)}{44},\ \
\beta_1=\ds\frac{\sqrt{135}}{22}+\ds\frac{i}{2}$

\bigskip

\begin{eqnarray*} \rm{center} & = &
\ds\frac{-(8\sqrt{135}+168)-2\sqrt{135}+22i}{-(8\sqrt{135}+168)-4\sqrt{135}}\\
& = & \ds\frac{-10\sqrt{135}-168+22i}{-12\sqrt{135}-168}\\ & \sim
& \un{0.9244-0.0716i} \end{eqnarray*}

\bigskip

$\un{k=1}$ $\gamma_2=-2\sqrt{15}-12,\ \
\beta_2=\ds\frac{\sqrt{15}}{2}+\ds\frac{i}{2}$

\bigskip

\begin{eqnarray*} \rm{center} & = &
\ds\frac{-2(2\sqrt{15}+12)-(\sqrt{15}-i)}{-2(2\sqrt{15}+12)-2\sqrt{15}}\\
& = & \ds\frac{-5\sqrt{15}-24+i}{-6\sqrt{15}-24}\\ & \sim &
\un{0.9180-0.0212i} \end{eqnarray*}

\bigskip

\begin{eqnarray*} m(z_0)=\ds\frac{z_0}{z_0+1} & = &
\ds\frac{\left(2+\ds\frac{\sqrt{135}}{4}\right)
\left(3+\ds\frac{\sqrt{135}}{4}\right)+\ds\frac{9}{16}-\ds\frac{3}{4}i}
{\left(3+\ds\frac{\sqrt{135}}{4}\right)^2+\ds\frac{9}{10}}\\ &
\sim & 0.8333-0.0212i \end{eqnarray*}

\bigskip

slope of radius of $m$ (tangent line to $A_1$): -0.5532

slope of tangent to $m$ (tangent line to $A_1$): 1.8075

angle of tangent to $\bf{R}$: $\phi_1=1.0655$

\newpage
slope of radius of $m$ (tangent line to $A_1$): $\sim 0$

slope of tangent to $m$ (tangent line to $A_1$): $\sim \infty$

angle of tangent to $\bf{R}$: $\phi_2=1.57-8$

and so the angle between $m(A_2)$ and $m(A_1)$ at $z_0$ is

$$\phi_2-\phi_1=0.5053$$ (as desired)
\end{document}
